Mathematics I Pharmaceutical Engineering MATH-102 Lecture Notes PDF
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German International University - GIU
2024
Dr. Karim Kamal
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These lecture notes cover topics in mathematics, specifically differentiation, including the chain rule, trigonometric and hyperbolic functions, and applications like implicit and logarithmic differentiation. The notes also include examples and exercises.
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Mathematics I Pharmaceutical Engineering MATH-102 Lecture 10 Dr. Karim Kamal In-Class Assignment (2) ❖ When? In the week starting 7/12/2024. ❖ Where? In your tutorials ❖ What? It will cover Lectures 8,9 + Worksheet 6. Dr. Karim Kamal 2 Lecture 10 – Outli...
Mathematics I Pharmaceutical Engineering MATH-102 Lecture 10 Dr. Karim Kamal In-Class Assignment (2) ❖ When? In the week starting 7/12/2024. ❖ Where? In your tutorials ❖ What? It will cover Lectures 8,9 + Worksheet 6. Dr. Karim Kamal 2 Lecture 10 – Outline ❖ Chain Rule ❖ Differentiation of Trigonometric and Hyperbolic Functions ❖ Chain Rule in Trigonometric and Hyperbolic Functions Dr. Karim Kamal 3 Rules of Differentiation Rule 9 Chain Rule If there are 2 differentiable functions 𝑓(𝑥) and 𝑔(𝑥): 1. If we define 𝐹 𝑥 = ( 𝑓 ∘ 𝑔) (𝑥), then the derivative of 𝐹(𝑥) is 𝐹′(𝑥) = 𝑓 ′(𝑔(𝑥)) 𝑔′ (𝑥) 2. If we have 𝑦 = 𝑓(𝑢) and 𝑢 = 𝑔(𝑥), then the derivative of 𝑦 is 𝑑𝑦 𝑑𝑦 𝑑𝑢 =. 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑦 where is evaluated at 𝑢 = 𝑔(𝑥). 𝑑𝑢 Dr. Karim Kamal 4 Exercise 3 ❖ Find 𝑦′ if: 1) 𝑦 = 𝑥 2 + 4𝑥 3 2) 𝑦 = 5𝑥 4 + 2𝑥 2 + 5 5 3) 𝑦 = 𝑒 3𝑥 + 3𝑥 − 1 2 4) 𝑦 = 𝑥 3 + 5𝑥 5 5) 𝑦 = 6 𝑒 2𝑥 Dr. Karim Kamal 5 Exercise 3 – Sol. ❖ Find 𝑦′ if: 1) 𝑦 = 𝑥 2 + 4𝑥 3 𝑦 ′ = 3 𝑥 2 + 4𝑥 2. (2𝑥 + 4) 2) 𝑦 = 5𝑥 4 + 2𝑥 2 + 5 5 𝑦 ′ = 5 5𝑥 4 + 2𝑥 2 + 5 4. (20𝑥 3 + 4𝑥) 3) 𝑦 = 𝑒 3𝑥 + 3𝑥 − 1 2 𝑦 ′ = 2 𝑒 3𝑥 + 3𝑥 − 1. (3𝑒 3𝑥 + 3) 4) 𝑦 = 𝑥 3 + 5𝑥 = 𝑥 3 + 5𝑥 1/2 ′ 1 3 − 1 3𝑥 2 + 5 𝑦 = 𝑥 + 5𝑥 2. 3𝑥 2 + 5 = 2 2 𝑥 3 + 5𝑥 Dr. Karim Kamal 6 Exercise 3 – Sol. ❖ Find 𝑦′ if: 2𝑥 5 5) 𝑦 = 6 𝑒 2𝑥 =6𝑒 5 2 2𝑥 12 2𝑥 12 5 𝑦′ =6 𝑒5 = 𝑒5 = 𝑒 2𝑥 5 5 5 Dr. Karim Kamal 7 Exercise 4 𝑑𝑦 ❖ Find given the following functions: 𝑑𝑥 1) 𝑦 = 𝑒 3𝑥 ln(2𝑥 + 1) 𝑥 𝑥 2 +7𝑥 2) 𝑦 = 𝑒 +2 10 3) 𝑦 = 1 + 𝑥 + 1 Dr. Karim Kamal 8 Exercise 4 – Sol. 𝑑𝑦 ❖ Find given the following functions: 𝑑𝑥 1) 𝑦 = 𝑒 3𝑥 ln(2𝑥 + 1) 3𝑥 𝑑𝑦 1 2𝑒 = 𝑒 3𝑥.. 2 + 3𝑒 3𝑥 ln(2𝑥 + 1) = + 3𝑒 3𝑥 ln(2𝑥 + 1) 𝑑𝑥 2𝑥 + 1 2𝑥 + 1 𝑥 2 +7𝑥 1/2 2 +7𝑥 2) 𝑦 = 𝑒 + 2𝑥 = 𝑒𝑥 + 2𝑥 𝑑𝑦 1 −1/2 𝑥 1/2 𝑥 2 +7𝑥 1 𝑥 𝑥 2 +7𝑥 = 𝑥 𝑒 + 2𝑥 + 7 2 ln 2 = 𝑒 + 2𝑥 + 7 2 ln 2 𝑑𝑥 2 2 𝑥 10 1/2 10 3) 𝑦 = 1 + 𝑥 + 1 = 1+ 𝑥+1 𝑑𝑦 1 9 1 1 5 9 −2 = 10 1 + 𝑥 + 1 2. 𝑥+1 = 1+ 𝑥+1 𝑑𝑥 2 𝑥+1 Dr. Karim Kamal 9 Exercise 5 𝑑𝑦 ❖ Find given the following functions: 𝑑𝑥 1) 𝑦 = ln(𝑥 2 − 6) 2) 𝑦 = 1 + ln(𝑥) 2 3) 𝑦 = log 2 𝑥 + 2 𝑥 4) 𝑦 = log 𝑥+1 Dr. Karim Kamal 10 Exercise 5 – Sol. 𝑑𝑦 ❖ Find given the following functions: 𝑑𝑥 1) 𝑦 = ln(𝑥 2 − 6) 𝑑𝑦 1 2𝑥 = 2. 2𝑥 = 2 𝑑𝑥 𝑥 − 6 𝑥 −6 2 2 1/2 2) 𝑦 = 1 + ln(𝑥) = 1 + ln(𝑥) 𝑑𝑦 1 2 −1/2. 2 1 ln 𝑥 = 1 + ln(𝑥) (ln 𝑥) = 𝑑𝑥 2 𝑥 𝑥 1 + ln(𝑥) 2 1/2 3) 𝑦 = log 2 𝑥 + 2 = log 2 𝑥 + 2 𝑑𝑦 1 1 1 −1/2 1 =.. 1/2 ln 2 2 𝑥+2 = 𝑑𝑥 𝑥+2 2(𝑥 + 2) ln 2 Dr. Karim Kamal 11 Exercise 5 – Sol. 𝑑𝑦 ❖ Find given the following functions: 𝑑𝑥 𝑥 4) 𝑦 = log 𝑥+1 𝑑𝑦 (𝑥 + 1) 1 𝑥 + 1 − 𝑥 1 =.. 2 = 𝑑𝑥 𝑥 ln 10 𝑥 + 1 𝑥 𝑥 + 1 ln 10 Another Solution: 𝑦 = log 𝑥 − log(𝑥 + 1) 𝑑𝑦 1 1 𝑥+1−𝑥 1 = − = = 𝑑𝑥 𝑥 ln 10 𝑥 + 1 ln 10 𝑥 𝑥 + 1 ln 10 𝑥 𝑥 + 1 ln 10 Dr. Karim Kamal 12 Derivatives of Trigonometric Functions Dr. Karim Kamal 13 Example ❖ Prove that: 𝑑 tan 𝑥 = sec 2 𝑥 𝑑𝑥 Dr. Karim Kamal 14 Example – Sol. ❖ Prove that: 𝑑 tan 𝑥 = sec 2 𝑥 𝑑𝑥 let 𝑦 = tan 𝑥 sin 𝑥 𝑦= cos 𝑥 𝑑𝑦 cos 𝑥 cos 𝑥 − − sin 𝑥 sin 𝑥 = 𝑑𝑥 cos 2 𝑥 𝑑𝑦 cos 2 𝑥 + sin2 𝑥 1 2𝑥 = = = sec 𝑑𝑥 cos 2 𝑥 cos 2 𝑥 Dr. Karim Kamal 15 Exercise ❖ Find 𝑦′ if: 1. 𝑦 = 𝑥 + tan 𝑥 2. 𝑦 = 𝑥 2 sin 𝑥 3. 𝑦 = sin 𝑥 /(1 + cos 𝑥) Dr. Karim Kamal 16 Exercise – Sol. 𝑦 = 𝑥 + tan 𝑥 𝑦 = 𝑥 1/2 + tan 𝑥 1 −1 𝑦 = 𝑥 2 + sec 2 𝑥 ′ 2 ′ 1 𝑦 = + sec 2 𝑥 2 𝑥 Dr. Karim Kamal 17 Exercise – Sol. 𝑦 = 𝑥 2 sin 𝑥 𝑦 ′ = 2𝑥. sin 𝑥 + 𝑥 2 cos 𝑥 Dr. Karim Kamal 18 Exercise – Sol. sin 𝑥 𝑦= 1 + cos 𝑥 cos 𝑥 1 + cos 𝑥 − − sin 𝑥 sin 𝑥 𝑦′ = 1 + cos 𝑥 2 cos 𝑥 + cos 2 𝑥 + sin2 𝑥 𝑦′ = 1 + cos 𝑥 2 ′ cos 𝑥 + 1 𝑦 = 1 + cos 𝑥 2 ′ 1 𝑦 = 1 + cos 𝑥 Dr. Karim Kamal 19 Derivatives of Trig./Hyperbolic Functions Dr. Karim Kamal 20 Example ❖ Prove that: 𝑑 sinh 𝑥 = cosh 𝑥 𝑑𝑥 Dr. Karim Kamal 21 Example – Sol. ❖ Prove that: 𝑑 sinh 𝑥 = cosh 𝑥 𝑑𝑥 let 𝑦 = sinh 𝑥 𝑒 𝑥 − 𝑒 −𝑥 𝑦= 2 1 𝑥 𝑦 = (𝑒 − 𝑒 −𝑥 ) 2 𝑑𝑦 1 𝑥 = (𝑒 − −1 𝑒 −𝑥 ) 𝑑𝑥 2 𝑥 + 𝑒 −𝑥 𝑑𝑦 1 𝑥 𝑒 = (𝑒 + 𝑒 −𝑥 ) = = cosh 𝑥 𝑑𝑥 2 2 Dr. Karim Kamal 22 Summary: Using the Chain Rule Dr. Karim Kamal 23 Exercise ❖ Find 𝑑𝑦/𝑑𝑥 for: 1. 𝑦 = 4 cos 𝑥 3 2. 𝑦 = 𝑥 3 + csc 𝑥 3. 𝑦 = sin( 1 + cos 2𝑥) 4. 𝑦 = tan(4𝑥 3 + 𝑥) Dr. Karim Kamal 24 Exercise – Sol. 𝑦 = 4 cos 𝑥 3 𝑑𝑦 = 4 − sin 𝑥 3. (3𝑥 2 ) 𝑑𝑥 𝑑𝑦 = −12𝑥 2 sin 𝑥 3 𝑑𝑥 Dr. Karim Kamal 25 Exercise – Sol. 𝑦 = 𝑥 3 + csc 𝑥 𝑦 = 𝑥 3 + csc 𝑥 1/2 𝑑𝑦 1 3 − 1 = 𝑥 + csc 𝑥 2. (3𝑥 2 − csc 𝑥 cot 𝑥) 𝑑𝑥 2 𝑑𝑦 3𝑥 2 − csc 𝑥 cot 𝑥 = 𝑑𝑥 2 𝑥 3 + csc 𝑥 Dr. Karim Kamal 26 Exercise – Sol. 𝑦 = sin( 1 + cos 2𝑥) 𝑦 = sin 1 + cos 2𝑥 1/2 1 1 𝑑𝑦 1 − = cos 1 + cos 2𝑥 2. 1 + cos 2𝑥 2. (− sin 2𝑥)(2) 𝑑𝑥 2 𝑑𝑦 cos 1 + cos 2𝑥. (sin 2𝑥) =− 𝑑𝑥 1 + cos 2𝑥 Dr. Karim Kamal 27 Exercise – Sol. 𝑦 = tan(4𝑥 3 + 𝑥) 𝑑𝑦 = sec 2 4𝑥 3 + 𝑥. (12𝑥 2 + 1) 𝑑𝑥 𝑑𝑦 = 12𝑥 2 + 1. sec 2 4𝑥 3 + 𝑥 𝑑𝑥 Dr. Karim Kamal 28 Exercise ❖ Find 𝑑𝑦/𝑑𝑥 for: 1. 𝑦 = 𝑒 𝑥 sinh 𝑥 2. 𝑦 = tanh 5𝑥 − 7 sech3 2𝑥 3. 𝑦 = cosh 𝑒 𝑥 + 3csch 𝑥 Dr. Karim Kamal 29 Exercise - Sol. 𝑦 = 𝑒 𝑥 sinh 𝑥 𝑑𝑦 = 𝑒 𝑥 sinh 𝑥 + 𝑒 𝑥 cosh 𝑥 𝑑𝑥 𝑑𝑦 = 𝑒 𝑥 (sinh 𝑥 + cosh 𝑥) 𝑑𝑥 Dr. Karim Kamal 30 Exercise - Sol. 𝑦 = tanh 5𝑥 − 7 sech3 2𝑥 𝑑𝑦 = sech2 5𝑥 5 − 7 × 3(sech2 2𝑥)(− sech 2𝑥 tanh 2𝑥)(2) 𝑑𝑥 𝑑𝑦 = 5 sech2 5𝑥 + 42 sech3 2𝑥 tanh 2𝑥 𝑑𝑥 Dr. Karim Kamal 31 Exercise - Sol. 𝑦 = cosh 𝑒 𝑥 + 3csch 𝑥 𝑑𝑦 = sinh 𝑒 𝑥 𝑒 𝑥 + − csch 𝑥 coth 𝑥 3csch 𝑥 ln 3 𝑑𝑥 𝑑𝑦 = 𝑒 𝑥 sinh 𝑒 𝑥 − 3csch 𝑥 ln 3 (csch 𝑥 coth 𝑥) 𝑑𝑥 Dr. Karim Kamal 32 Thank You ☺ Dr. Karim Kamal 33 Mathematics I Pharmaceutical Engineering MATH-102 Lecture 11 Dr. Karim Kamal Quiz (2) ❖ When? Sunday 22/12/2024, 2nd slot. ❖ What? It will cover Lectures 10,11 &12 + Worksheets 7 &8. ❖ Where? At S1.107 Dr. Karim Kamal 2 Lecture 11 – Outline ❖ Applications of differentiation ❑ Implicit Differentiation ❑ Logarithmic Differentiation ❑ Relative extrema and critical points of a function First derivative test Dr. Karim Kamal 3 Implicit Differentiation: Dr. Karim Kamal 4 Example (1) 𝒅𝒚 𝒅𝒚 (1) Find 𝒚 = ′ if : 𝒙 + 𝒚 − 𝟗𝒙𝒚 = 𝟎 𝟑 𝟑 (2) Find 𝒚 = ′ if : 𝒚 + 𝐥𝐧 𝒙𝒚 = 𝟏 𝒅𝒙 𝒅𝒙 Solution: Solution: Dr. Karim Kamal 5 Example (2) Dr. Karim Kamal 6 Example (3) Find 𝒚′ if 𝒙𝟑 + 𝒚𝟑 = 𝟔𝒙𝒚. Hence, find the equation of the tangent line to the curve at the point (𝟑, 𝟑). Solution: Differentiating both sides with respect to 𝒙. 3 𝑥 2 + 3 𝑦 2 𝑦′ = 6 𝑥 𝑦′ + 6 𝑦 Solving for 𝒚′. 𝑦 2 𝑦′ − 2 𝑥 𝑦′ = 2 𝑦 − 𝑥 2 𝑦′ 𝑦 2 − 2 𝑥 = 2 𝑦 − 𝑥 2 2 𝑦 − 𝑥2 ∴ 𝑦′ = 2 𝑦 −2𝑥 At the point 3,3 ⇒ 𝑦′ = −1 𝑦−3 The equation of tangent: = −1 ⇒ 𝑦 = 6 − 𝑥 𝑥−3 Dr. Karim Kamal 7 Logarithmic Differentiation: Case 1 ❖ If we have a complicated expression with multiple products and quotients, where applying the standard rules would be a complicated process, then 1. Take the natural logarithm (ln) of both sides. 2. Simplify the resulting expression (using the rules of logarithms). 3. Differentiate implicitly. Dr. Karim Kamal 8 Example ❖ Find 𝑦′ for the following function: 3 𝑥2 − 8 𝑥3 + 1 𝑦= 𝑥 6 − 7𝑥 + 1 Dr. Karim Kamal 9 Example – Sol. 3 𝑥2 − 8 𝑥3 + 1 𝑦= 𝑥 6 − 7𝑥 + 1 Apply natural logarithm ln on both sides: 3 𝑥2 − 8 𝑥3 + 1 ln 𝑦 = ln 𝑥 6 − 7𝑥 + 1 3 ln 𝑦 = ln 𝑥 2 − 8 + ln 𝑥 3 + 1 − ln(𝑥 6 − 7𝑥 + 1) ln 𝑦 = ln 𝑥 2 − 8 1/3 + ln 𝑥 3 + 1 1/2 − ln(𝑥 6 − 7𝑥 + 1) 1 1 ln 𝑦 = ln(𝑥 2 − 8) + ln 𝑥 3 + 1 − ln(𝑥 6 − 7𝑥 + 1) 3 2 Implicitly differentiate both sides with respect to (w.r.t.) 𝑥: 1 𝑑𝑦 1 2𝑥 1 3𝑥 2 6𝑥 5 − 7 = + − 6 𝑦 𝑑𝑥 3 (𝑥 2 − 8) 2 𝑥 3 + 1 (𝑥 − 7𝑥 + 1) 𝑑𝑦 1 2𝑥 1 3𝑥 2 6𝑥 5 − 7 =𝑦 + − 6 𝑑𝑥 3 𝑥2 − 8 2 𝑥3 + 1 𝑥 − 7𝑥 + 1 3 𝑑𝑦 𝑥 2 − 8 𝑥 3 + 1 1 2𝑥 1 3𝑥 2 6𝑥 5 − 7 = + − 6 𝑑𝑥 𝑥 6 − 7𝑥 + 1 3 𝑥2 − 8 2 𝑥3 + 1 𝑥 − 7𝑥 + 1 Dr. Karim Kamal 10 Example Dr. Karim Kamal 11 Exercise 1 ❖ Find 𝑦′ for the following functions: 3 5 1) 𝑦 = 𝑥 𝑥 + 3 3𝑥 − 1 𝑥 3/4 𝑥 2 +1 2) 𝑦 = 3𝑥+2 5 3 𝑥 2 −3 3) 𝑦 = 1+𝑥 5 Dr. Karim Kamal 12 Exercise 1 – Sol. 1 3 5 𝑦 = 𝑥 𝑥 + 3 3𝑥 − 1 3 5 ln 𝑦 = ln( 𝑥 𝑥 + 3 3𝑥 − 1 ) 3 5 ln 𝑦 = ln 𝑥 + ln 𝑥 + 3 + ln 3𝑥 − 1 ln 𝑦 = ln 𝑥 1/2 + ln 𝑥 + 3 1/3 + ln 3𝑥 − 1 1/5 1 1 1 ln 𝑦 = ln 𝑥 + ln(𝑥 + 3) + ln(3𝑥 − 1) 2 3 5 Differentiate both sides w.r.t. 𝑥 1 𝑑𝑦 1 1 1 1 1 3 = + + 𝑦 𝑑𝑥 2 𝑥 3 (𝑥 + 3) 5 (3𝑥 − 1) 𝑑𝑦 11 1 1 1 3 =𝑦 + + 𝑑𝑥 2𝑥 3 𝑥 + 3 5 3𝑥 − 1 𝑑𝑦 3 5 1 1 3 = 𝑥 𝑥 + 3 3𝑥 − 1 + + 𝑑𝑥 2𝑥 3𝑥 + 9 15𝑥 − 5 Dr. Karim Kamal 13 Exercise 1 – Sol. 2 𝑥 3/4 𝑥 2 + 1 𝑦= 3𝑥 + 2 5 3 𝑥4 𝑥2 + 1 ln 𝑦 = ln 3𝑥 + 2 5 3 5 ln 𝑦 = ln 𝑥 4 + ln 𝑥 2 + 1 − ln 3𝑥 + 2 3 ln 𝑦 = ln 𝑥 4 + ln 𝑥 2 + 1 1/2 − ln 3𝑥 + 2 5 3 1 ln 𝑦 = ln 𝑥 + ln(𝑥 2 + 1) − 5 ln(3𝑥 + 2) 4 2 Differentiate both sides w.r.t. 𝑥 1 𝑑𝑦 3 1 1 2𝑥 3 = + − 5 𝑦 𝑑𝑥 4 𝑥 2 (𝑥 2 + 1) (3𝑥 + 2) 𝑑𝑦 3 1 1 2𝑥 3 =𝑦 + − 5 𝑑𝑥 4 𝑥 2 𝑥2 + 1 3𝑥 + 2 𝑑𝑦 𝑥 3/4 𝑥 2 + 1 3 𝑥 15 = + − 𝑑𝑥 3𝑥 + 2 5 4𝑥 𝑥2 + 1 3𝑥 + 2 Dr. Karim Kamal 14 Exercise 1 – Sol. 3 3 𝑥2 − 3 𝑦= 1 + 𝑥5 3 𝑥2 − 3 ln 𝑦 = ln 1 + 𝑥5 1/3 𝑥2 − 3 ln 𝑦 = ln 1 + 𝑥5 1 𝑥2 − 3 ln 𝑦 = ln 3 1 + 𝑥5 1 ln 𝑦 = ln 𝑥 2 − 3 − ln 1 + 𝑥 5 3 1 1 ln 𝑦 = ln 𝑥 2 − 3 − ln 1 + 𝑥 5 3 3 Differentiate both sides w.r.t. 𝑥: 1 𝑑𝑦 1 2𝑥 1 5𝑥 4 = − 𝑦 𝑑𝑥 3 (𝑥 2 − 3) 3 1 + 𝑥 5 𝑑𝑦 1 2𝑥 1 5𝑥 4 =𝑦 − 𝑑𝑥 3 𝑥2 − 3 3 1 + 𝑥5 𝑑𝑦 3 𝑥 2 − 3 1 2𝑥 1 5𝑥 4 = − 𝑑𝑥 1 + 𝑥5 3 𝑥2 − 3 3 1 + 𝑥5 Dr. Karim Kamal 15 Logarithmic Differentiation: Case 2 ❖ If we want to differentiate a function of 𝑥 that is raised to another function of 𝑥, i.e. 𝑔(𝑥) 𝑦 = 𝑓 𝑥 1. Take the natural logarithm of both sides. 2. Simplify the resulting expression (using the rules of logarithms). 3. Apply implicit differentiation. Dr. Karim Kamal 16 Example ❖ Find 𝑦′ for the following function: 𝑥 𝑦=𝑥 Dr. Karim Kamal 17 Example – Sol. 𝑦=𝑥 𝑥 ln 𝑦 = ln 𝑥 𝑥 ln 𝑦 = 𝑥 ln 𝑥 ln 𝑦 = 𝑥 1/2 ln 𝑥 Differentiate both sides w.r.t. 𝑥: 1 𝑑𝑦 1 −1/2 1 = 𝑥 ln 𝑥 + 𝑥 1/2 𝑦 𝑑𝑥 2 𝑥 1 𝑑𝑦 1 −1/2 = 𝑥 ln 𝑥 + 𝑥 −1/2 𝑦 𝑑𝑥 2 1 𝑑𝑦 1 1 = ln 𝑥 + 𝑦 𝑑𝑥 2 𝑥 𝑥 𝑑𝑦 1 1 =𝑦 ln 𝑥 + 𝑑𝑥 2 𝑥 𝑥 𝑑𝑦 1 1 =𝑥 𝑥 ln 𝑥 + 𝑑𝑥 2 𝑥 𝑥 Dr. Karim Kamal 18 Exercise 2 ❖ Find 𝑦′ for the following functions: 1) 𝑦 = 1 + 𝑥 1/𝑥 2) 𝑦 = 𝑥 3 − 2𝑥 ln 𝑥 Dr. Karim Kamal 19 Exercise 2 – Sol. 1 𝑦 = 1 + 𝑥 1/𝑥 ln 𝑦 = ln 1 + 𝑥 1/𝑥 1 ln 𝑦 = ln(1 + 𝑥) 𝑥 ln 𝑦 = 𝑥 −1 ln(1 + 𝑥) Differentiate both sides w.r.t. 𝑥: 1 𝑑𝑦 −2 1 = −𝑥 ln 1 + 𝑥 + 𝑥 −1 𝑦 𝑑𝑥 (1 + 𝑥) 𝑑𝑦 1 = 𝑦 −𝑥 −2 ln 1 + 𝑥 + 𝑥 −1 𝑑𝑥 1+𝑥 𝑑𝑦 1/𝑥 ln 1 + 𝑥 1 = 1+𝑥 − 2 + 𝑑𝑥 𝑥 𝑥 1+𝑥 Dr. Karim Kamal 20 Exercise 2 – Sol. 2 𝑦 = 𝑥 3 − 2𝑥 ln 𝑥 ln 𝑦 = ln 𝑥 3 − 2𝑥 ln 𝑥 ln 𝑦 = ln 𝑥 ln 𝑥 3 − 2𝑥 Differentiate both sides w.r.t. 𝑥: 1 𝑑𝑦 1 3𝑥 2−2 = ln 𝑥 3 − 2𝑥 + 3 ln 𝑥 𝑦 𝑑𝑥 𝑥 𝑥 − 2𝑥 𝑑𝑦 1 3𝑥 2−2 = 𝑦 ln 𝑥 3 − 2𝑥 + 3 ln 𝑥 𝑑𝑥 𝑥 𝑥 − 2𝑥 𝑑𝑦 ln 𝑥 3 − 2𝑥 3𝑥 2 − 2 = 𝑥 3 − 2𝑥 ln 𝑥 + 3 ln 𝑥 𝑑𝑥 𝑥 𝑥 − 2𝑥 Dr. Karim Kamal 21 Local Maxima and Minima Source: Weir, Hass, Giordano, Thomas’ Calculus, 11th Ed. ❖ A function 𝑓 has a local maximum value at an interior point 𝑐 of its domain if 𝑓 𝑐 ≥ 𝑓 (𝑥) for all 𝑥 in some open interval containing 𝑐. ❖ A function 𝑓 has a local minimum value at an interior point 𝑒 of its domain if 𝑓 𝑒 ≤ 𝑓 (𝑥) for all 𝑥 in some open interval containing 𝑒. Dr. Karim Kamal 22 Relative (Local) Extrema and Critical Points of a Function Dr. Karim Kamal 23 Critical Points of a Function ❖ The first derivative theorem for local extreme values states that if a function 𝑓 has a local maximum or minimum value at an interior point 𝑐 of its domain, and if 𝑓′ is defined at 𝑐, then 𝑓 ′(𝑐) = 0 ❖ An interior point of the domain of a function 𝑓 where 𝑓′ is zero or undefined is a critical point of 𝑓. ❖ How to find the critical point of a function? 1. Find 𝑓′. 2. If 𝑓′ exists and is not a quotient function, then just put 𝑓′ = 0 and solve for 𝑥. 3. If 𝑓′ exists and is a quotient function, then just let both the numerator of 𝑓′ and its denominator = 0 and solve for 𝑥. Dr. Karim Kamal 24 Critical Points of a Function ❖ Critical points are the locations on a graph where the derivative equals zero or is undefined. ❖ Critical points are candidates for the location of maxima and minima of a function. ❖ Functions exhibit interesting behavior at their critical points. Dr. Karim Kamal 25 Critical Points of a Function ❖ The geometric meaning of critical points of a function is that at such points, the tangent line to the function is 1. horizontal 2. vertical 3. does not exist (In case the function has no derivative at this point) Dr. Karim Kamal 26 Corollary for the First Derivative Test for Monotonic Functions Source: Weir, Hass, Giordano, Thomas’ Calculus, 11th Ed. The function 𝑓(𝑥) = 𝑥 2 is monotonic on the intervals ] − ∞, 0] and [0, ∞[, but it is not monotonic on ] − ∞, ∞[. The function 𝑓(𝑥) = 𝑥 3 − 12 𝑥 − 5 is monotonic on 3 separate intervals. ❖ Suppose that f is continuous on [𝑎, 𝑏] and differentiable on ]𝑎, 𝑏[ ❖ If 𝑓′(𝑥) > 0 at each point 𝑥 ∈ ]𝑎, 𝑏[, then 𝑓 is increasing on [𝑎, 𝑏] ❖ If 𝑓′(𝑥) < 0 at each point 𝑥 ∈ ]𝑎, 𝑏[, then 𝑓 is decreasing on [𝑎, 𝑏] Dr. Karim Kamal 27 The First Derivative Test for Local Extrema ❖ Suppose that 𝑐 is a critical point of a continuous function 𝑓, and that 𝑓 is differentiable at every point in some interval containing 𝑐 except possibly at 𝑐 itself. Moving across 𝑐 from left to right, 1) If 𝑓′ changes from -ve to +ve at 𝑐, then 𝑓 has a local minimum at c; 2) If 𝑓′ changes from +ve to -ve at 𝑐, then 𝑓 has a local maximum at 𝑐; 3) If 𝑓′ does not change sign at 𝑐 (that is, 𝑓′ is positive on both sides of 𝑐 or negative on both sides), then 𝑓 has no local extremum at 𝑐. Dr. Karim Kamal 28 Example 1 ❖ Find the critical points of 𝑓(𝑥) = 𝑥 3 − 12 𝑥 − 5 and identify the intervals on which 𝑓 is increasing and decreasing. Dr. Karim Kamal 29 Example 1 – Sol. ❖ The function 𝑓 is everywhere continuous and differentiable. The first derivative is obtained as follows: 𝑓(𝑥) = 𝑥 3 − 12 𝑥 − 5 𝑓′(𝑥) = 3 𝑥 2 − 12 ❖ Put 𝑓 ′ 𝑥 = 0 3𝑥 2 − 12 = 0 𝑥2 − 4 = 0 𝑥2 = 4 𝑥2 = 4 𝑥 =2 𝑥 = ±2 𝑥 = −2, 2 ∈ 𝐷𝑓 → 𝑇ℎ𝑒𝑦 𝑎𝑟𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑓 Dr. Karim Kamal 30 Example 1 – Sol. ❖ These critical points subdivide the domain of 𝑓 into intervals ] − ∞, −2[ , ] − 2, 2[ and ]2, ∞[ on which 𝑓′ is either positive or negative. ❖ We determine the sign of 𝑓′ by evaluating 𝑓′ at a convenient point in each subinterval. ❖ The behavior of 𝑓 is determined then by applying the Corollary to each subinterval. Source: Weir, Hass, Giordano, Thomas’ Calculus, 11th Ed. Dr. Karim Kamal 31 Example 1 – Sol. A plot of the of the function 𝑓(𝑥) = 𝑥 3 − 12 𝑥 − 5. Source: Weir, Hass, Giordano, Thomas’ Calculus, 11th Ed. Dr. Karim Kamal 32 Exercise 1 ❖ Find the points of local maximum and minimum values for the function 𝑓 (𝑥) = 𝑥 1 − 𝑥 2/5 Dr. Karim Kamal 33 Exercise 1 – Sol. 𝑓 𝑥 = 𝑥 1 − 𝑥 2/5 ′ 2/5 2 − 3 𝑓 𝑥 = 1−𝑥 + 1 − 𝑥 5 −1 𝑥 5 ′ 2/5 2𝑥 5 1 − 𝑥 − 2𝑥 5 − 5𝑥 − 2𝑥 5 − 7𝑥 𝑓 𝑥 = 1−𝑥 − 3 = 3 = 3 = 3 5 1−𝑥 5 5 1−𝑥 5 5 1−𝑥 5 5 1−𝑥 5 5 ❖ Put the numerator of 𝑓′(𝑥) equal 0: 5 − 7𝑥 = 0 → 7𝑥 = 5 → 𝑥 = ∈ 𝐷𝑓 7 3 3 ❖ Put the denominator of 𝑓′ 𝑥 equal 0: 5 1 − 𝑥 5 =0 → 1−𝑥 5 =0 → 1 − 𝑥 = 0 → 𝑥 = 1 ∈ 𝐷𝑓 5 ❖ 𝑓 ′ 𝑥 = 0 at 𝑥 = and 𝑓′(𝑥) DNE at 𝑥 = 1 7 5 ❖ 𝑥 = 1, 𝑥 = are critical points of the function 𝑓(𝑥) 7 Dr. Karim Kamal 34 Exercise 1 – Sol. 5−7𝑥 ❖ 𝑓′ 𝑥 = 3 5 1−𝑥 5 5 ❖ The critical points 𝑥 = 1 and 𝑥 = divide the real line into 3 intervals: 7 5 5 −∞, , ,1 and 1, ∞ 7 7 Intervals 5 5 1, ∞ −∞, ,1 7 7 6 𝑓′ evaluated 𝑓′ 0 = 1 𝑓′ = −0.6428 𝑓 ′ 2 = 1.8 7 Sign of 𝑓′ +ve −ve +ve Behaviour of 𝑓 Increasing Decreasing Increasing Dr. Karim Kamal 35 Exercise 1 – Sol. 5 ❖ Therefore, the point 𝑥 = is a 7 local maximum. ❖ The point 𝑥 = 1 is a local minimum. ❖ By substituting in the function: 2/5 ❖ 𝑓 (𝑥) = 𝑥 1 − 𝑥 5 ❖ Local max. at approx. , 0.433 7 ❖ Local min. at (1,0) Dr. Karim Kamal 36 Thank You ☺ Dr. Karim Kamal 37 Mathematics I Pharmaceutical Engineering MATH-102 Lecture 12 Dr. Karim Kamal Quiz (2) ❖ When? Sunday 22/12/2024, 2nd slot. ❖ What? It will cover Lectures 10,11 &12 + Worksheets 7 &8. ❖ Where? At S1.107 Dr. Karim Kamal 2 Lecture 12 – Outline ❖ Relative extrema and critical points of a function Second derivative test ❖ L’Hopital’s Rule Dr. Karim Kamal 3 The Second Derivative Test for Local Extrema ❖ Suppose 𝑓′′ is continuous on an open interval that contains 𝑥 = 𝑐, such that 𝑐 ∈ 𝐷𝑓 1. If 𝑓 ′(𝑐) = 0 and 𝑓 ′′(𝑐) < 0, then 𝑓 has a local maximum at 𝑥 = 𝑐 2. If 𝑓 ′(𝑐) = 0 and 𝑓 ′′(𝑐) > 0, then 𝑓 has a local minimum at 𝑥 = 𝑐 3. If 𝑓 ′(𝑐) = 0 and 𝑓 ′′(𝑐) = 0, then the test fails. The function 𝑓 may have a local maximum, a local minimum, or neither. Dr. Karim Kamal 4 Example 2 ❖ Find the points of local maximum and minimum values for the function 𝑓 (𝑥) = 2 𝑥 3 − 3 𝑥 2 − 12 𝑥, using the second derivative test Dr. Karim Kamal 5 Example 2 – Sol. 𝑓 (𝑥) = 2 𝑥 3 − 3 𝑥 2 − 12 𝑥 𝑓 ′ 𝑥 = 6𝑥 2 − 6𝑥 − 12 𝑓 ′′ 𝑥 = 12𝑥 − 6 ❖ Put 𝑓 ′ 𝑥 = 0: 6𝑥 2 − 6𝑥 − 12 = 0 → 𝑥 2 − 𝑥 − 2 = 0 → (𝑥 + 1)(𝑥 − 2) = 0 𝑥 = −1 𝑜𝑟 𝑥 = 2 𝑥 = −1, 2 ∈ 𝐷𝑓 → 𝑡ℎ𝑒𝑦 𝑎𝑟𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑓(𝑥) ❖ Check 𝑓 ′′ 𝑥 at these critical points: 𝑓 ′′ −1 = −18 < 0 → 𝑙𝑜𝑐𝑎𝑙 𝑚𝑎𝑥. 𝑓 ′′ 2 = 18 > 0 → 𝑙𝑜𝑐𝑎𝑙 𝑚𝑖𝑛. Dr. Karim Kamal 6 Example 2 – Sol. ❖ By substituting in the function itself 𝑓 (𝑥) = 2 𝑥 3 − 3 𝑥 2 − 12 𝑥 ❖ Local max. at (−1,7) ❖ Local min. at (2, −20) Dr. Karim Kamal 7 Exercise 2 ❖ The function 𝑓 (𝑥) = 2 𝑥 3 − 15 𝑥 2 + 24 𝑥 − 7 has critical points 𝑥 = 1 and 4 ❖ Use the second derivative test for extrema to determine whether 𝑓(1) and 𝑓(4) are maxima, minima or the test fails. Dr. Karim Kamal 8 Exercise 2 – Sol. 𝑓 (𝑥) = 2 𝑥 3 − 15 𝑥 2 + 24 𝑥 − 7 𝑓 ′ 𝑥 = 6𝑥 2 − 30𝑥 + 24 𝑓 ′′ 𝑥 = 12𝑥 − 30 ❖ Check 𝑓′′(𝑥) at the critical points 𝑥 = 1 and 4: 𝑓 ′′ 1 = −18 < 0 → 𝑙𝑜𝑐𝑎𝑙 𝑚𝑎𝑥. 𝑓 ′′ 4 = 18 > 0 → 𝑙𝑜𝑐𝑎𝑙 𝑚𝑖𝑛. Dr. Karim Kamal 9 Exercise 2 – Sol. ❖ Substituting in the function with the critical points: 𝑓 (𝑥) = 2 𝑥 3 − 15 𝑥 2 + 24 𝑥 − 7 ❖ Local max. at (1,4) ❖ Local min. at (4, −23) Dr. Karim Kamal 10 Indeterminate Forms and L’Hopital’s Rule ❖ Limits that result in 0 ∞ , , 0 × ∞, ∞ − ∞, 00 , ∞0 , 1∞ 0 ∞ are called indeterminate forms and can be classified into: 0 ±∞ 1. Standard: and 0 ±∞ 2. Non–standard: 0 × ∞, ∞ − ∞, 00 , ∞0 , 1∞ Note that: ∞ + ∞ = ∞ and ∞ × ∞ = ∞ ❖ Sometimes, it is easy to use the so called L’Hopital’s rule to find the values of these limits. ❖ In this course, we will only be concerned with the standard case. Dr. Karim Kamal 11 L’Hopital’s Rule ❖ Suppose that 𝑓(𝑎) = 𝑔(𝑎) = 0 or that 𝑓(𝑎) = 𝑔(𝑎) = ∞. Also suppose 𝑓 and 𝑔 are differentiable on an open interval 𝐼 containing 𝑎, and that 𝑔′(𝑥) ≠ 0 on 𝐼 if 𝑥 ≠ 𝑎. Then 𝑓 𝑥 𝑓′ 𝑥 lim = lim 𝑥→𝑎 𝑔(𝑥) 𝑥→𝑎 𝑔′(𝑥) assuming that the limit on the right side exists. Guillaume De l'Hôpital 1661 - 1704 Dr. Karim Kamal 12 How to use L’Hopital’s Rule? ❖ To find 𝑓 𝑥 lim 𝑥→𝑎 𝑔(𝑥) ❖ by L’Hopitals rule, continue to differentiate 𝑓 and 𝑔 as many times needed, as long as we are getting 0/0 or ∞/∞ at 𝑥 = 𝑎. ❖ Keep in mind that L’Hopital’s rule does not apply when either the numerator or denominator has a finite non–zero limit. Dr. Karim Kamal 13 Examples for the Standard Case of 0/0 ❖ Evaluate each of the following limits: 𝑥 2 −4 1. lim 𝑥→2 𝑥−2 sin 𝑥 2. lim 𝑥→0 𝑥 1−sin 𝑥 3. lim 𝑥→𝜋/2 cos 𝑥 tan 𝑥 4. lim 𝑥→0− 𝑥 2 1−cos 𝑥 5. lim 𝑥2 𝑥→0 𝑥 −4/3 6. lim 𝑥→∞ sin 1Τ𝑥 Dr. Karim Kamal 14 Examples for the Standard Case of 0/0 𝑥2 − 4 lim 𝑥→2 𝑥 − 2 𝑥2 − 4 0 lim = Direct Substitution 𝑥→2 𝑥 − 2 0 𝑥2 − 4 𝐿 2𝑥 lim ֜ lim =2 2 =4 𝑥→2 𝑥 − 2 𝑥→2 1 𝐿 Note that: the symbol “֜” signifies applying L'Hopital's rule. Dr. Karim Kamal 15 Examples for the Standard Case of 0/0 sin 𝑥 lim 𝑥→0 𝑥 sin 𝑥 0 lim = Direct Substitution 𝑥→0 𝑥 0 sin 𝑥 𝐿 cos 𝑥 lim ֜ lim =1 𝑥→0 𝑥 𝑥→0 1 Dr. Karim Kamal 16 Examples for the Standard Case of 0/0 1 − sin 𝑥 lim 𝑥→𝜋/2 cos 𝑥 1 − sin 𝑥 1 − 1 0 lim = = Direct Substitution 𝑥→𝜋/2 cos 𝑥 0 0 1 − sin 𝑥 𝐿 − cos 𝑥 cos 𝑥 0 lim ֜ lim = lim = =0 𝑥→𝜋/2 cos 𝑥 𝑥→𝜋/2 − sin 𝑥 𝑥→𝜋/2 sin 𝑥 1 Dr. Karim Kamal 17 Examples for the Standard Case of 0/0 tan 𝑥 lim 𝑥→0− 𝑥 2 tan 𝑥 0 lim = (Direct Substitution) 𝑥→0− 𝑥 2 0 tan 𝑥 𝐿 sec 2 𝑥 lim ֜ lim− = −∞ 𝑥→0− 𝑥 2 𝑥→0 2𝑥 The limit doesn't exit. Dr. Karim Kamal 18 Examples for the Standard Case of 0/0 1 − cos 𝑥 lim 𝑥→0 𝑥2 1 − cos 𝑥 1 − 1 0 lim 2 = = Direct Substitution 𝑥→0 𝑥 0 0 1 − cos 𝑥 𝐿 −(− sin 𝑥) sin 𝑥 0 lim 2 ֜ lim = lim = Direct Substitution 𝑥→0 𝑥 𝑥→0 2𝑥 𝑥→0 2𝑥 0 sin 𝑥 𝐿 cos 𝑥 1 lim ֜ lim = 𝑥→0 2𝑥 𝑥→0 2 2 Note that: you can apply L’Hôpital’s rule as many times as necessary as long as the fraction is still indeterminate Dr. Karim Kamal 19 Examples for the Standard Case of 0/0 𝑥 −4/3 lim 𝑥→∞ sin 1Τ𝑥 𝑥 −4/3 0 lim = Direct Substitution 𝑥→∞ sin 1Τ𝑥 0 4 −7/3 4 −7/3 𝑥 −4/3 𝐿 − 𝑥 − 𝑥 lim ֜ lim 3 = lim 3 𝑥→∞ sin 1Τ𝑥 𝑥→∞ (cos 1Τ𝑥 )(−1/𝑥 2 ) 𝑥→∞ −𝑥 −2 cos 1Τ𝑥 4 −1/3 𝑥 0 = lim 3 = =0 𝑥→∞ cos 1Τ𝑥 1 Dr. Karim Kamal 20 Examples for the Standard Case of ∞/∞ ❖ Evaluate each of the following limits: 𝑥 1. lim 𝑥→∞ 𝑒 𝑥 ln 𝑥 2. lim+ csc 𝑥 𝑥→0 1−ln 𝑥 3. lim+ 𝑥→0 𝑒 1/𝑥 Dr. Karim Kamal 21 Examples for the Standard Case of ∞/∞ 𝑥 lim 𝑥 𝑥→∞ 𝑒 𝑥 ∞ lim = (Direct Substitution) 𝑥→∞ 𝑒 𝑥 ∞ 𝑥 𝐿 1 lim ֜ lim 𝑥 = 0 𝑥→∞ 𝑒 𝑥 𝑥→∞ 𝑒 Dr. Karim Kamal 22 Examples for the Standard Case of ∞/∞ ln 𝑥 𝑦 = csc 𝑥 lim 𝑥→0+ csc 𝑥 ln 𝑥 −∞ lim = 𝑥→0+ csc 𝑥 ∞ 1 ln 𝑥 𝐿 𝑥 lim+ ֜ lim+ 𝑥→0 csc 𝑥 𝑥→0 − csc 𝑥 cot 𝑥 1 𝑥 − sin 𝑥. tan 𝑥 = lim+ = lim+ 𝑥→0 1 1 𝑥→0 𝑥 −. sin 𝑥 tan 𝑥 sin 𝑥 = − lim+ lim+ tan 𝑥 = − 1 0 = 0 𝑥→0 𝑥 𝑥→0 Dr. Karim Kamal 23 Examples for the Standard Case of ∞/∞ 1 − ln 𝑥 lim 𝑥→0+ 𝑒 1/𝑥 1 − ln 𝑥 1 − (−∞) ∞ lim+ 1/𝑥 = = 𝑥→0 𝑒 ∞ ∞ 1 1 − ln 𝑥 𝐿 − 1 lim ֜ lim+ 𝑥 = lim 𝑥→0+ 𝑒 1/𝑥 𝑥→0 1 1/𝑥 𝑥→0+ 1 1/𝑥 − 2𝑒 𝑒 𝑥 𝑥 1 = lim+ 𝑥. lim+ 1/𝑥 = 0 0 = 0 𝑥→0 𝑥→0 𝑒 Dr. Karim Kamal 24 Exercises: ❖ Evaluate each of the following limits: 𝑥2 − 9 𝑒𝑥 − 1 1. lim 4. lim 𝑥→3 𝑥 − 3 𝑥→0 𝑥 3 𝑥2 + 3 𝑥 − 4 3𝑥 − 5𝑥 2. lim 5 5. lim 𝑥→1 𝑥 − 5 𝑥 + 4 𝑥→0 𝑥 ln 𝑥 𝑥 3. lim 1+𝑥−1− 𝑥→1 𝑥 − 1 6. lim 2 𝑥→0 𝑥2 Final Answers: 1. 6 4. ∞ 2. ∞ 5. ln(3/5) 3. 1 6. −1/8 Dr. Karim Kamal 25 Thank You ☺ Dr. Karim Kamal 26 Mathematics I Pharmaceutical Engineering MATH-102 Lecture 1 Dr. Karim Kamal Course Information - Instructor ❖ Course Instructor: Dr. Karim Kamal ❖ Office location: S1.304 ❖ Email address: [email protected] N.B.: Please clearly state your name, ID, the course you take, brief question or inquiry, no request to break rules. ❖ Office Hours: By appointment through email. ❖ Textbooks: R. N. Greenwell, N. P. Ritchey and M. L. Lial, Calculus for the Life Sciences. S. Waner and S. R. Costenoble, Finite Mathematics and Applied Calculus. G. B. Thomas, Jr., M. D. Weir, J. Hass, F. R. Giordano, Thomas’ Calculus, 11th ed., Pearson, Addison Wesley, 2005. Dr. Karim Kamal 2 Course Information – Team ❖ TA: Mr. Mohamed Khalefa ❖ Office location: S1.212 ❖ Email address: [email protected] Dr. Karim Kamal 3 Course Assessment ❖ 10% Assignments ❖ 20% Quizzes ❖ 30% Midterm ❖ 40% Final exam Dr. Karim Kamal 4 Why do we as Pharmaceutical Engineering students, need Math? Dr. Karim Kamal 5 Why do Pharmaceutical Engineers, study Mathematics ? ❖ Describe various physiological processes using mathematical equations. ❖ Develop Mathematical Models that can predict and get a good estimate of various scientific phenomena. ❖ Statistics: collect some sample data and report for example the rate of spread of a specific disease or pandemic (for e.g. COVID-19). Dr. Karim Kamal 6 Course Outline ❖ Algebra and pre-calculus prerequisites ❖ Limits and derivatives of functions ❖ Optimization of functions ❖ Definite and Indefinite Integration ❖ Linear Algebra ❖ Complex Numbers ❖ Applications in pharmaceutical engineering Dr. Karim Kamal 7 Lecture 1 - Outline 1. Sets ▪ Relations between Sets ▪ Set Operations 2. The real numbers ▪ Real Numbers and the Real Line ▪ Order Properties of Real Numbers 3. Exponents and Radicals (part 1) ▪ Radicals ▪ Powers of Real Numbers Dr. Karim Kamal 8 A set is a group of objects ❖ All modern mathematical analysis and its daily life applications (management, economics, social sciences, etc...) is build on the theory of sets. In everyday life, we often need to group together "objects" that are alike in some way. In this case we have a collection of objects that form what we call a "set" because they have something in common. ❖ Example The letters a, e, i, o and u can be grouped together to form a set. ❑ The objects: letters a, e, i, o and u ❑ The common property: vowels in English alphabet ❑ The set: V is the set of all vowels in English alphabet Dr. Karim Kamal 9 Basic set theory (1/3) Dr. Karim Kamal 10 Basic set theory (2/3) Dr. Karim Kamal 11 Basic set theory (3/3) Dr. Karim Kamal 12 Classification of Numbers ❖ Natural Numbers: ℕ ❖ Integers: ℤ ❖ Rational Numbers: ℚ ❖ Irrational Numbers: ℚ′ ❖ Real Numbers: ℝ Dr. Karim Kamal 13 Classification of Numbers ❖ Natural Numbers: ℕ = {1,2,3,4, … … … } ❖ Integers: ℤ = {… … … , −3, −2, −1, 0, 1, 2, 3, … … … } ℤ contains Positive Integers ℤ+ (Natural Numbers ℕ), Negative Integers ℤ− and Zero ❖ Rational Numbers: ℚ = {𝑥 = 𝑎/𝑏; 𝑎, 𝑏 ∈ ℤ, 𝑏 ≠ 0} 2 −4 4 4 100 57 Examples: , =− = , , = 57 3 9 9 −9 9 1 Rational Numbers can be also represented in their decimal notation: 1 o Either ending in infinite sequence of zeros: 4 = 0.25000000 … = 0.25 o Or repeating sequence of numbers: 1 3 23 = 0.333333 … = 0. ത, 3 = 0.428571428571 … = 0. 428571, = 2.09090909 … = 2. 09 3 7 11 Dr. Karim Kamal 14 Classification of Numbers (Cont’d) ❖ Irrational Numbers: ℚ′ The set of numbers that can not be expressed in the form of a fraction 𝑎/𝑏 (or in other words: their decimal representation does not contain repeating sequence of numbers nor infinite sequence of zeros) are called irrational numbers. 3 𝜋 = 3.14159265 …, 𝑒 = 2.718281828459 …, 2 = 1.41421356 …, 5 = 1.70997594 … ❖ Real Numbers: ℝ ℚ ∪ ℚ′ = ℝ = (−∞, ∞) The union of the Rational numbers ℚ and the Irrational numbers ℚ′ is the set of Real Numbers ℝ. ℚ ∩ ℚ′ = 𝜙 The intersection of the Rational numbers ℚ and the Irrational numbers ℚ′ is an empty set. N.B.: ℕ, ℤ contain discrete data values, while ℝ contains continuous data values. Dr. Karim Kamal 15 Venn Diagram – Classification of Numbers Real Numbers ℝ Rational Numbers ℚ Irrational Numbers ℚ′ Repeating Decimals 𝑎/𝑏 2 Integers ℤ 𝜋 0 −2 𝑒 Natural Num. ℕ −3 2 1.211212 … 1 −1 3 Dr. Karim Kamal 16 The Real Number Line ❖ Natural Numbers ℕ ❖ Integer Numbers ℤ ❖ Rational Numbers ℚ − 5 − 2 2 5 𝑒 𝜋 ❖ Irrational Numbers ℚ′ − 5 − 2 2 5 𝑒 𝜋 ❖ Real Numbers ℝ Dr. Karim Kamal 17 Geometric Representation of Real Numbers on the Number Line ❖ Every real number is represented by a unique point on a horizontal straight line called the number line. ❖ Conversely, every point on the number line represents one and only one real number. ❖ We say that 𝑎 < 𝑏 if 𝑏 − 𝑎 is a positive real number, or equivalently, if 𝑎 precedes 𝑏 on the number line ❖ 𝑎 ≤ 𝑏 if 𝑎 < 𝑏 or 𝑎 = 𝑏 Dr. Karim Kamal 18 Questions – Rational or Irrational? 1) Is 0.999999999 … = 0. 9ത a rational or irrational number ? Rational, because after the decimal point there are infinite repeating sequence of 9’s. 2) Is 0.999999999 … = 0. 9ത = 1 ? Let 𝑝 = 0.999999999 … 10𝑝 = 9.999999999 … 10𝑝 − 𝑝 = 9.999999999 … − 0.999999999 … 9𝑝 = 9 𝑝=1 ∴ 0.999999999 … = 1 Dr. Karim Kamal 19 Questions – Represent 𝟓 on the Number Line 3) Show how can you represent 5 on the number line ? A First, let’s revise Pythagoras’ Theorem: In any right-angled triangle ABC, right-angled at B, 𝑦 ℎ If the side lengths are 𝑥, 𝑦 and ℎ where ℎ is the hypotenuse. Therefore, 𝑥 2 + 𝑦 2 = ℎ2 B C 𝑥 Dr. Karim Kamal 20 Questions – Represent 𝟓 on the Number Line 3) Show how can you represent 5 on the number line ? A 𝐿𝑒𝑡 𝑥 = 5 𝑥2 = 5 𝑥2 + 1 6 𝑦 ℎ ℎ= = =3 2 2 𝑥2 − 1 4 𝑦= = =2 2 2 B C 𝑥 𝑁. 𝐵. : 𝑥 2 + 𝑦 2 = ℎ2 → ℎ2 − 𝑦 2 = 𝑥 2 32 − 22 = 9 − 4 = 5 = 𝑥 2 Dr. Karim Kamal 21 Questions – Represent 𝟓 on the Number Line 3) Show how can you represent 5 on the number line ? 𝑦 = 2 𝑐𝑚 ℎ = 3 𝑐𝑚 Consider the vertex point B is 𝑥 = 5 𝑐𝑚 the origin point on the number line, measure 𝑦 = 2𝑐𝑚 vertically, then adjust your compass to be A opened ℎ = 3𝑐𝑚 and put its pin on the point A, then try to draw an arc that intersects the horizontal line (the Number line) 𝑦=2 ℎ=3 at the point C (which is exactly at a distance ℎ = 3𝑐𝑚 from A and at a distance 𝑥 = 5 𝑐𝑚 from the point B). B 0 𝑥= 5 5 C Dr. Karim Kamal 22 Sum or Difference of Two Numbers 4) Is the sum or difference of any two integers also an integer? Yes. Examples: 5 + 9 = 14 3 + 100 = 103 20 − 3 = 17 4 − 30 = −26 5) Is the sum or difference of any two rational numbers also a rational number? 𝑎 𝑐 𝑎𝑑 ± 𝑐𝑏 Yes. General Example: Let 𝑎, 𝑏, 𝑐, 𝑑 be integers, then 𝑏 ± 𝑑 = 𝑏𝑑 Since the multiplication of two integers also an integer, and the sum or difference of two integers also an integer, therefore, 𝑎𝑑 ± 𝑐𝑏 is a rational number. 𝑏𝑑 6) Is the sum or difference of any two real numbers also a real number? Yes. Dr. Karim Kamal 23 Sum or Difference of Two Numbers 7) Is the sum or difference of any two irrational numbers also an irrational number? Not always true. Examples: 𝜋−𝜋=0 2+ 2=2 2 Rational Irrational Dr. Karim Kamal 24 Integer Exponents of Real Numbers ❖ If 𝑎 is any real number and 𝑛 is any positive integer, then we define 𝑎𝑛 = 𝑎. 𝑎. 𝑎. 𝑎. 𝑎 …. 𝑎 → 𝑛 times 𝑎0 = 1, 𝑎≠0 1 𝑎−𝑛 = 𝑛, 𝑎≠0 𝑎 0𝑛 = 0, 𝑛≠0 00 → undefined 0−𝑛 → undefined Dr. Karim Kamal 25 Laws of Exponents ❖ If 𝑎, 𝑏 are real numbers and 𝑛, 𝑚 are integers, 𝑎𝑛 𝑎𝑚 = 𝑎𝑛+𝑚 𝑎𝑚 𝑚−𝑛 , = 𝑎 𝑎≠0 𝑎𝑛 𝑎𝑛 𝑚 = 𝑎𝑛𝑚 (𝑎𝑏)𝑛 = 𝑎𝑛 𝑏 𝑛 𝑎 𝑛 𝑎𝑛 = 𝑛, 𝑏≠0 𝑏 𝑏 Dr. Karim Kamal 26 Integer Exponents – Examples “1” ❖ 52 = 5 × 5 = 25 ❖ 24 = 2 × 2 × 2 × 2 = 16 ❖ −10 3 = (−10) × (−10) × (−10) = −1000 ❖ −10 2 = −10 × −10 = 100 1 1 1 ❖ 10−2 = = = 102 10×10 100 1 1 ❖ 𝑥 −1 = = 𝑥1 𝑥 ❖ 032 = 0 ❖ (1,000,000)0 = 1 ❖ −5𝑥𝑦 2 0 = 1, 𝑥, 𝑦 ≠ 0 Dr. Karim Kamal 27 Integer Exponents – Examples “2” ❖ 23 22 = 23+2 = 25 = 32 ❖ 32 Τ34 = 32−4 = 3−2 = 1Τ32 = 1Τ9 ❖ 4/3 3 = 43 Τ33 = 64/27 ❖ 𝑥 3 𝑥 −4 = 𝑥 3−4 = 𝑥 −1 = 1/𝑥 ❖ 𝑥 3 /𝑥 −2 = 𝑥 3+2 = 𝑥 5 ❖ −2𝑦 4 = −2 4 𝑦 4 = 16𝑦 4 Dr. Karim Kamal 28 Integer Exponents - Exercise ❖ Write the following expression using only positive exponents: 𝑥 4 𝑦 −1 3 𝑓= 𝑥 −2 𝑦 5 Dr. Karim Kamal 29 Integer Exponents - Exercise ❖ Write the following expression using only positive exponents: 𝑥 4 𝑦 −1 3 𝑓= 𝑥 −2 𝑦 5 Solution: 𝑥 4 3 𝑦 −1 3 𝑓= 𝑥 −2 𝑦 5 𝑥 12 𝑦 −3 𝑥 14 𝑓 = −2 5 = 𝑥 12+2 𝑦 −3−5 = 𝑥 14 𝑦 −8 = 8 𝑥 𝑦 𝑦 Dr. Karim Kamal 30 Thank You ☺ Dr. Karim Kamal 31 Mathematics I Pharmaceutical Engineering MATH-102 Lecture 2 Dr. Karim Kamal Lecture 2 - Outline ❖ Radicals and Rational Exponents ❖ Solving Equations ❖ Solving Equations including absolute values Dr. Karim Kamal 2 Radicals and Rational Exponents ❖ If 𝑎 is a non-negative (positive or zero) real number and 𝑚, 𝑛 are positive integers, 𝑚, 𝑛 > 1, then we define 𝑛 𝑎 = 𝑎1/𝑛 , the previous relation also holds if 𝑎 is negative real and 𝑛 is odd integer 𝑚/𝑛 𝑛 𝑛 𝑚 𝑎 = 𝑎𝑚 = 𝑎 , where 𝑛, 𝑚 do not have a common factor Dr. Karim Kamal 3 Radicals and Rational Exponents ❖ Laws of Radicals: 𝑎 is a positive real number 𝑛 𝑎 ∈ ℝ if ቊ 𝑎 is a negative real number and 𝑛 is odd In other words: If 𝑛 is even integer, 𝑎 must be positive real number Dr. Karim Kamal 4 Radicals and Rational Exponents ❖ Laws of Radicals: If 𝑎, 𝑏 are real numbers, 𝑛, 𝑚 are positive integers, 𝑛, 𝑚 > 1 𝑛 𝑎𝑛 = 𝑎 if 𝑛 is even 𝑛 𝑎𝑛 = 𝑎 if 𝑛 is odd 𝑚 𝑛 𝑎= 𝑚𝑛 𝑎 = 𝑎1/(𝑚𝑛) , if 𝑚 or 𝑛 or both are even , then 𝑎 must be positive real 𝑛 𝑛 𝑎𝑏 = 𝑎 × 𝑏 = 𝑎1/𝑛 𝑏1/𝑛 𝑛 𝑛 𝑎 𝑛 𝑎 𝑎1/𝑛 = 𝑛 = 1/𝑛 , 𝑏≠0 𝑏 𝑏 𝑏 Dr. Karim Kamal 5 Radicals and Rational Exponents - Examples ❖ 𝑎 = 𝑎1/2 , 𝑎≥0 ❖ 3 𝑎 = 𝑎1/3 , 𝑎 is any real number ❖ −4 is undefined in ℝ!! ❖ − 4 = −2 3 ❖ −8 = −2 ❖ 0=0 3 ❖ 0=0 3 3 ❖ 02/3 = 02 = 0=0 Dr. Karim Kamal 6 Radicals and Rational Exponents - Examples 3 2 3 2 ❖ 82/3 = 8 = 23 = 22 = 4 3 ❖ 41.5 = 43/2 = 4 = 23 = 8 5 1 3 N.B. 1.5 = 1 = 1 = 10 2 2 Dr. Karim Kamal 7 Radicals and Rational Exponents - Examples 3 3 ❖ 52/3 = 52 = 25 ≈ 2.924 ❖ −8 3/2 = −8 3 = −512 is undefined in ℝ!! 2/3 3 3 ❖ −8 = −8 2 = 64 = 4 5 4 4 ❖ −5 4 = −5 5 = −3125 is undefined in ℝ!! ❖ Common Mistake: 2/10 10 2 −32 = −32 =2 Wrong ‼ 10 2 −32 2/10 = −32 is undefined in ℝ Wrong ‼ 2/10 1/5 5 5 −32 = −32 = −32 = −2 5 = −2 Correct ✓ Dr. Karim Kamal 8 Radicals and Rational Exponents - Exercise ❖ Simplify the following expression: 3 𝑎2 𝑎 2 𝑎−5 𝑓= 𝑎5 Dr. Karim Kamal 9 Radicals and Rational Exponents - Exercise ❖ Simplify the following expression: 3 𝑎2 𝑎 2 𝑎−5 𝑓= 𝑎5 Solution: 3 3 2 1/2 2 5 2 𝑎 𝑎 𝑎−5 𝑎2 𝑎−5 𝑎15 𝑎−5 𝑎10 𝑓= = = = = 𝑎 5 𝑎5 𝑎5 𝑎5 𝑎5 Dr. Karim Kamal 10 Important Formulas ❖ Difference Between Two Squares: 𝐴2 − 𝐵2 = 𝐴 − 𝐵 𝐴 + 𝐵 ❖ Binomial Theorem (when the power of the Binomial is 2): 𝐴 + 𝐵 2 = 𝐴2 + 2𝐴𝐵 + 𝐵2 𝐴 − 𝐵 2 = 𝐴2 − 2𝐴𝐵 + 𝐵2 Dr. Karim Kamal 11 Division by Zero ❖ A common mistake: 2 =0 Wrong ‼ 0 2 =2 Wrong ‼ 0 2 is undefined Correct ✓ 0 ❖ However: 0 =0 2 0 =0 −300 Dr. Karim Kamal 12 Division by Zero - Example 1) Let us define 𝑎=𝑏 2) Multiply both sides by 𝑎 𝑎2 = 𝑎𝑏 3) Subtract 𝑏 2 from both sides 𝑎2 − 𝑏 2 = 𝑎𝑏 − 𝑏 2 4) Factorize both sides 𝑎 − 𝑏 𝑎 + 𝑏 = 𝑏(𝑎 − 𝑏) 5) Divide both sides by (𝑎 − 𝑏) 𝑎+𝑏 =𝑏 6) Substitute 𝑏 = 𝑎 2𝑏 = 𝑏 7) Divide both sides by 𝑏 2=1 Wrong ‼ We made a mistake in step (5), since 𝑎 = 𝑏, therefore 𝑎 − 𝑏 = 0. This means we divided by 0 !!!! Thus, it resulted in a wrong answer. Dr. Karim Kamal 13 Techniques of Solving Equations and Inequalities ❖ Introduction to Equations ❖ Solving linear equations ❖ Solving quadratic equations ❖ Solving other types of equations Dr. Karim Kamal 14 Introduction to Equations ❖ Q1. What is an equation? ❖ A1. It is a statement in which 2 algebraic expressions are equal ❖ Q2. What is the solution of an equation? ❖ A2. It is the set of values of the variable(s) which make the equation a true statement ❖ Example 1: 2𝑥 + 4𝑥 = 6𝑥 → 6𝑥 = 6𝑥 → 𝑥=𝑥 This equation is true for all values of x (identity equation) ❖ Example 2: 2𝑥 + 1 = 5 → 2𝑥 = 4 → 𝑥=2 This equation is true for some values of x (conditional equation) Dr. Karim Kamal 15 Solving Linear Equations ❖ The standard form of a linear equation is 𝑎𝑥 + 𝑏 = 𝑐 ❖ It has only one solution, in the form 𝑐−𝑏 𝑥 = 𝑎 ❖ Example: Solve the equation 2𝑥 + 1 = 7 ❖ Solution: 2𝑥 + 1 = 7 → 2𝑥 = 6 → 𝑥 = 3 Dr. Karim Kamal 16 Solving Quadratic Equations ❖ The standard form of a quadratic equation is 𝑎 𝑥2 + 𝑏 𝑥 + 𝑐 = 0 ❖ It has 2 solutions, obtained by one of the following methods 1) Factorization 2) The standard quadratic formula Dr. Karim Kamal 17 Solving Quadratic Equations: Factorization ❖ Example: Solve the equation 𝑥 2 − 7 𝑥 + 12 = 0 ❖ Solution: 𝑥 2 − 7𝑥 + 12 = 0 (𝑥 − 4) (𝑥 − 3) = 0 Which means that (𝑥 − 4) = 0 → 𝑥 = 4 Or (𝑥 − 3) = 0 → 𝑥 = 3 ❖ Finally, this equation has 2 solutions 𝑥 = 4 or 𝑥 = 3 and its solution set is 𝑥 ∈ {3, 4} Dr. Karim Kamal 18 Solving Quadratic Equations: Factorization ❖ Example: Solve the equation 𝑥 2 − 𝑥 − 12 = 0 ❖ Solution: 𝑥 2 − 𝑥 − 12 = 0 (𝑥 − 4) (𝑥 + 3) = 0 ❖ Which means that (𝑥 − 4) = 0 → 𝑥 = 4 Or (𝑥 + 3) = 0 → 𝑥 = −3 ❖ Finally, this equation has 2 solutions 𝑥 = 4 or 𝑥 = −3 and its solution set is 𝑥 ∈ {−3, 4} Dr. Karim Kamal 19 Solving Quadratic Equations: The Standard Quadratic Formula −𝑏 ± 𝑏2 − 4 𝑎 𝑐 𝑥1,2 = 2𝑎 ❖ Example: Solve the equation 𝑥 2 − 𝑥 − 12 = 0 ❖ Solution: We first identify that 𝑎 = 1, 𝑏 = −1 and 𝑐 = −12, then we plug these values into the above formula 1 ± −1 2 − 4 (1) (−12) 1 ± 1 + 48 1 ± 7 𝑥1,2 = = = 2(1) 2 2 ❖ 𝑥1 = 4 and 𝑥2 = −3 ❖ Thus, this equation has 2 solutions 𝑥 = 4 or 𝑥 = −3 and its solution set is 𝑥 ∈ {−3, 4} Dr. Karim Kamal 20 Solving Quadratic Equations - Exercise ❖ Solve the following equation: 𝑥 − 2 2 = 12 ❖ Solution: 𝑥 2 − 4𝑥 + 4 = 12 𝑥 2 − 4𝑥 + 4 − 12 = 0 𝑥 2 − 4𝑥 − 8 = 0 𝑎 = 1, 𝑏 = −4, 𝑐 = −8 4± −4 2 − 4(1)(−8) 4 ± 16 + 32 4 ± 48 4 ± 4 3 𝑥1,2 = = = = 2(1) 2 2 2 𝑥1 = 2 + 2 3 𝑥2 = 2 − 2 3 Dr. Karim Kamal 21 Equations Involving Absolute Values |𝑢| 1) If the equation takes the form |𝑢| = 𝑘 Then, depending on 𝑘, different solutions are reached as follows i. 𝑢 = ±𝑘, 𝑖𝑓 𝑘 > 0 ii. 𝑢 = 0, 𝑖𝑓 𝑘 = 0 iii. no solution, 𝑖𝑓 𝑘 < 0 2) If the equation takes the form |𝑢| = |𝑣|, then 𝑢 = ±𝑣 Dr. Karim Kamal 22 Solving Quadratic Equations - Exercise ❖ Solve the following equation: 2 𝑥 − 2 = 12 ❖ Another Solution: Taking of both sides: 𝑥−2 2 = 12 𝑥 − 2 = 12 𝑥 − 2 = ± 12 𝑥 = 2 ± 12 𝑥1,2 = 2 ± 2 3 𝑥 ∈ {2 − 2 3, 2 + 2 3} Dr. Karim Kamal 23 Equations Involving Absolute Values - Exercises 1) |4 𝑥 − 3| = 5 2) |2 𝑥 − 3| = |𝑥 − 6| 3) |𝑥 2 − 6 𝑥| = |12 − 7 𝑥| 6 4) 𝑥 − 1 = 64 5) 𝑥 + 5 3 = −27 Dr. Karim Kamal 24 Equations Involving Absolute Values – Exercises – Solutions (1) |4 𝑥 − 3| = 5 4𝑥 − 3 = ±5 4𝑥 − 3 = 5 Or 4𝑥 − 3 = −5 4𝑥 = 8 4𝑥 = −2 𝑥=2 𝑥 = −1/2 Solution Set is 𝑥 ∈ {−1/2,2} Dr. Karim Kamal 25 Equations Involving Absolute Values – Exercises – Solutions (2) |2 𝑥 − 3| = |𝑥 − 6| 2𝑥 − 3 = ± 𝑥 − 6 2𝑥 − 3 = 𝑥 − 6 Or 2𝑥 − 3 = −(𝑥 − 6) 2𝑥 − 𝑥 = −6 + 3 2𝑥 − 3 = −𝑥 + 6 𝑥 = −3 2𝑥 + 𝑥 = 6 + 3 3𝑥 = 9 𝑥=3 Solution Set is 𝑥 ∈ {−3,3} Dr. Karim Kamal 26 Equations Involving Absolute Values – Exercises – Solutions (3) 𝑥 2 − 6 𝑥 = 12 − 7 𝑥 𝑥 2 − 6𝑥 = ±(12 − 7𝑥) 𝑥 2 − 6𝑥 = 12 − 7𝑥 Or 𝑥 2 − 6𝑥 = −(12 − 7𝑥) 𝑥 2 − 6𝑥 + 7𝑥 = 12 𝑥 2 − 6𝑥 = −12 + 7𝑥 𝑥 2 + 𝑥 − 12 = 0 𝑥 2 − 6𝑥 − 7𝑥 + 12 = 0 𝑥+4 𝑥−3 =0 𝑥 2 − 13𝑥 + 12 = 0 𝑥 = −4 𝑜𝑟 𝑥 = 3 𝑥 − 1 𝑥 − 12 = 0 𝑥=1 𝑜𝑟 𝑥 = 12 Solution Set is 𝑥 ∈ {−4,1,3,12} Dr. Karim Kamal 27 Equations Involving Absolute Values – Exercises – Solutions (4) 𝑥 − 1 6 = 64 6 Take of both sides 6 6 6 𝑥 − 1 = 64 𝑥−1 =2 𝑥 − 1 = ±2 𝑥−1=2 Or 𝑥 − 1 = −2 𝑥=3 𝑥 = −1 Solution Set is 𝑥 ∈ {−1, 3} Dr. Karim Kamal 28 Equations Involving Absolute Values – Exercises – Solutions (5) 𝑥 + 5 3 = −27 3 Take of both sides 3 3 3 𝑥 + 5 = −27 𝑥 + 5 = −3 𝑛 𝑵. 𝑩. : 𝐍𝐨 need for 𝐚𝐛𝐬𝐨𝐥𝐮𝐭𝐞 𝐯𝐚𝐥𝐮𝐞 here, as the 𝒏 in the is 𝐨𝐝𝐝 number. 𝑥 = −8 Solution Set is 𝑥 ∈ {−8} Dr. Karim Kamal 29 Thank You ☺ Dr. Karim Kamal 30 Mathematics I Pharmaceutical Engineering MATH-102 Lecture 3 Dr. Karim Kamal Lecture 3 - Outline ❖ Solving Inequalities ❖ Solving Inequalities including absolute values ❖ Introduction to Functions (Domain and Range) Dr. Karim Kamal 2 Intervals ❖ Q. What is an interval? ❖ A. It is any portion of the number line ❖ An interval could be bounded or unbounded ❖ A bounded interval is one which corresponds to a line segment on the number line. ❖ An unbounded interval is one which corresponds to a ray on the number line. Dr. Karim Kamal 3 Types of Intervals Bounded =]𝑎, 𝑏[ = [𝑎, 𝑏[ =]𝑎, 𝑏] Unbounded =]𝑎, ∞[ = [𝑎, ∞[ =] − ∞, 𝑏[ =] − ∞, 𝑏] =] − ∞, ∞[ Dr. Karim Kamal 4 Source: Weir, Hass, Giordano, Thomas’ Calculus, 11th Ed. Inequalities ❖ Q1. What is an inequality? ❖ A1. It is a statement in which 2 (or more) algebraic expressions are related to each other by one (or more) of the symbols < , > , ≤ 𝑜𝑟 ≥ ❖ Q2. What is the solution of an inequality? ❖ A2. It is the set of all values (usually an interval) of the variable that makes the inequality a true statement (satisfies the inequality) Dr. Karim Kamal 5 Rules for Inequalities ❖ If a, b, and c are real numbers, then: 1. 𝑎 < 𝑏 ⟹ 𝑎+𝑐 𝑏𝑐 Special case: 𝑎 < 𝑏 and 𝑐 = −1 ⟹ −𝑎 > −𝑏 1 5. 𝑎 > 0 , >0 𝑎 1 1 6. If a and b are both positive or both negative, then 𝑎 < 𝑏 ⟹ > 𝑎 𝑏 Dr. Karim Kamal 6 Absolute Value - Revisited ❖ Absolute Value Properties: −𝑎 = 𝑎 |𝑎𝑏| = |𝑎||𝑏| 𝑎 |𝑎| = 𝑏 |𝑏| 𝑎 + 𝑏 ≤ 𝑎 + |𝑏|: Triangle Inequality Dr. Karim Kamal 7 Inequalities Involving Absolute Values |𝑢| 1. If the inequality takes the form |𝑢| < 𝑘 Then, change it to: −𝑘 < 𝑢 < 𝑘 2. If the inequality takes the form |𝑢| > 𝑘 Then, change it to: 𝑢 > 𝑘 or 𝑢 < −𝑘 3. If the inequality takes the form |𝑢| > 𝑢 This is equivalent to 𝑢 < 0 In summary: i. 𝑢2 = |𝑢| ii. 𝑢2 < 𝑘 ↔ 𝑢 < 𝑘↔ − 𝑘 𝑘 ↔ 𝑢 > 𝑘 ↔ 𝑢 > 𝑘 𝑜𝑟 𝑢 < − 𝑘 Dr. Karim Kamal 8 Inequalities Involving Absolute Values |𝑢| - Exercises ❖ Solve each of the following inequalities 1) |2 𝑥 − 3| < 5 2) |𝑥 2 − 3| > 1 𝑥−4 3) 𝑥−1 ≥ 2 Dr. Karim Kamal 9 Inequalities Involving Absolute Values |𝑢| - Exercises – Solutions (1) ❖ Solve each of the following inequalities |2 𝑥 − 3| < 5 −5 < 2𝑥 − 3 < 5 −2 < 2𝑥 < 8 −1 < 𝑥 < 4 The solution set is 𝑥 ∈ ] − 1,4[ Dr. Karim Kamal 10 Inequalities Involving Absolute Values |𝑢| - Exercises – Solutions (2) ❖ Solve each of the following inequalities |𝑥 2 − 3| > 1 𝑥 2 − 3 < −1 Or 𝑥2 − 3 > 1 𝑥2 < 2 𝑥2 > 4 Apply on both sides Apply on both sides 𝑥2 < 2 𝑥2 > 4 𝑥 < 2 𝑥 >2 − 2 2 The solution set is ] − ∞, −2[ ∪ ] − 2, 2[ ∪ ]2, ∞[ Dr. Karim Kamal 11 Inequalities Involving Absolute Values |𝑢| - Exercises – Solutions (3) 𝑥 − 4 ≥ 2, 𝑥≠1 𝑥 − 1 𝑥−4 ≥2 𝑥−1 Square Both sides: 𝑥 − 4 2 ≥ 4 𝑥 − 1 2 𝑥 2 − 8𝑥 + 16 ≥ 4 𝑥 2 − 2𝑥 + 1 𝑥 2 − 8𝑥 + 16 ≥ 4𝑥 2 − 8𝑥 + 4 −3𝑥 2 + 12 ≥ 0 Divide both sides by − 3, ∶ 𝑥2 − 4 ≤ 0 𝑥2 ≤ 4 𝑥2 ≤ 4 𝑥 ≤2 −2 ≤ 𝑥 ≤ 2 The solution set is −2,2 − {1} Dr. Karim Kamal 12 Functions A function from a set 𝐷 to a A diagram showing a function as a kind of machine. set 𝑌 is a rule that assigns a unique (single) element 𝑓(𝑥) ∈ 𝑌 to each element 𝑥 ∈ 𝐷. In other words, each element in 𝐷 has one and only one corresponding value in 𝑌. A function from a set 𝐷 to a set 𝑌 assigns a unique element of 𝑌 to each element in 𝐷. Source: Weir, Hass, Giordano, Thomas’ Calculus, 11th Ed. Dr. Karim Kamal 16 Functions’ Representations ❖ There are 4 possible ways to represent a function: 1) Verbally, by a description in words 2) Numerically, by a table of values 3) Visually, by a graph 4) Algebraically, by an explicit formula Dr. Karim Kamal 17 Functions: Numerical, Graphical Representations ❖ This table shows values of the world population in 1900. ❖ If we plot these values, we get a graph called a scatter plot. ❖ It is possible to find an expression for a function that approximates the set of points → mathematical model ❖ The figure at the bottom-right shows a mathematical model for population growth. Dr. Karim Kamal 18 Functions: Algebraic Representation 𝑓(𝑥) = 2𝑥 − 1 𝑓(𝑥) = 𝑥 2 Dr. Karim Kamal 19 Testing Realtions: Dr. Karim Kamal 20 Functions: Graphical Check ❖ By using the vertical line test (V.L.T.) → A relation 𝑓 is a function if and only if (iff) its graph intersects each vertical line at most once Source: Waner, Costenoble, Applied Calculus, 5th Ed. Dr. Karim Kamal 21 Functions: Algebraic Check Dr. Karim Kamal 22 Functions: Domain and Range ❖ We refer to 𝑥 as the independent variable and to 𝑦 as the dependent variable. ❖ A function 𝑓 is a rule of correspondence that associates with each object 𝑥 in one set, called the domain, a single value 𝑓(𝑥) from a second set. The set of all values obtained is called the range of the function. Varberg, Purcell, Rigdon, Calculus, 11th Ed ❖ To completely specify a function 𝑓, we must state not only the rule of correspondence, but also the domain of the function. Dr. Karim Kamal 23 Functions: Domain and Range ❖ General rules: 1) You cannot divide by zero (i.e. the denominator of a fraction cannot be a zero). 2) You cannot take the square root (nor even root) of a negative number. Source: Weir, Hass, Giordano, Thomas’ Calculus, 11th Ed. −∞, 0 ∪ (0, ∞) = ℝ − {0} N.B: Within this course, we will only obtain the range graphically! However, the domain can be either obtained algebraically Dr. Karim Kamal or graphically! 24 Summary of domains Dr. Karim Kamal 25 Thank You ☺ Dr. Karim Kamal 26 Mathematics I Pharmaceutical Engineering MATH-102 Lecture 4 Dr. Karim Kamal Lecture 4 - Outline ❖ Even and odd functions ❖ Basic Functions Dr. Karim Kamal 2 Even and Odd Functions Dr. Karim Kamal 3 Basic Functions 1. Polynomials 2. Nth root functions “Radicals” 3. Rational functions 4. Absolute value function “Modulus” 5. Exponential functions 6. Logarithmic functions Dr. Karim Kamal 4 Polynomial Functions ❖ A function 𝑃(𝑥) is called an nth degree polynomial if it is of the form 𝑃 𝑥 = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + … + 𝑎1 𝑥 + 𝑎0 where 𝑛 is a non–negative integer and the numbers 𝑎0 , 𝑎1 , … , 𝑎𝑛 are real constants called coefficients of the polynomial. ❖ Note: A function 𝑃(𝑥) is called a polynomial if all the powers of 𝑥 are non- negative integers. ❖ The domain of any polynomial is = ] − ∞, +∞[ = ℝ ❖ The highest power of 𝑥 appearing in the polynomial is said to be the degree of the polynomial ❖ For example, 𝑃(𝑥) = 3 𝑥 2 − 2 𝑥 + 4 is a polynomial of degree 2 (it is a quadratic polynomial) Dr. Karim Kamal 5 Constant Polynomial ❖ The simplest polynomial function has a degree of zero ❖ For example, 𝑃(𝑥) = 3 / 2 is a zero–degree polynomial Dr. Karim Kamal 6 Polynomial of Degree 1 → Linear Function Dr. Karim Kamal 7 Polynomial of Degree 2 → Quadratic Function Dr. Karim Kamal 8 Graphs of Some Polynomials ❖ Do you recognize the degree of each of these polynomials? Source: Weir, Hass, Giordano, Thomas’ Calculus, 11th Ed. Dr. Karim Kamal 9 Graphs of Special Polynomials ❖ Can you identify the domain and range from the graph ? Source: Weir, Hass, Giordano, Thomas’ Calculus, 11th Ed. Dr. Karim Kamal 10 Radical Functions “Nth Root Functions” Source: Weir, Hass, Giordano, Thomas’ Calculus, 11th Ed. Dr. Karim Kamal 11 Rational Functions ❖ The quotient of two polynomials is called a rational function 𝑓(𝑥) and has the following form 𝑃 𝑥 𝑓 𝑥 = 𝑄(𝑥) ❖ The domain of the rational function is ℝ − {𝑄(𝑥) = 0} ❖ Example: The domain of the rational function 3 𝑥2 − 1 𝑓 𝑥 = 𝑥−4 ❖ is ℝ − {4} = {𝑥 ∶ 𝑥 ∈ ℝ , 𝑥 ≠ 4}. Dr. Karim Kamal 12 Some Special Rational Functions ❖ 𝑓 (𝑥) = 1 / 𝑥 ❖ The graph of 𝑓 (𝑥) = 1/𝑥 𝑛 will have a similar shape when 𝑛 is odd Dr. Karim Kamal 13 Some Special Rational Functions ❖ 𝑓 𝑥 = 1Τ𝑥 2 ❖ The graph of 𝑓 (𝑥) = 1/𝑥 𝑛 will have a similar shape when 𝑛 is even Dr. Karim Kamal 14 The Absolute Value Function 𝑓 (𝑥) = |𝑥| ❖ The absolute value function is an expression of the form 𝑓 (𝑥) = 𝑎 |𝑥 − 𝑏| + 𝑐, where 𝑎, 𝑏 and 𝑐 are real numbers ❖ How do the functions 𝑦 = 𝑥 and 𝑦 = |𝑥| compare to one another? Dr. Karim Kamal 15 Exponential Functions ❖ For any positive real number 𝑎 and any real number 𝑥, we define the exponential function as 𝑓 (𝑥) = 𝑎 𝑥 ∶ 𝑎 > 0, 𝑎 ≠ 1 ❖ The domain of any exponential function 𝑎𝑢 equals the domain of its power 𝑢. Dr. Karim Kamal 16 Exercise ❖ For exponential functions as in the following form 𝑓 (𝑥) = 𝑎 𝑥 ∶ 𝑎 > 0, 𝑎 ≠ 1 ❖ What is the case when 𝑎 = 1? What about when 𝑎 = 0? ❖ Solution: 1. When 𝑎 = 1 𝑎 = 1 → 𝑓 (𝑥) = 1𝑥 = 1 2. When 𝑎 = 0 𝑎 = 0 → 𝑓 (𝑥) = 0𝑥 = 0, 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑡ℎ𝑎𝑡 𝑥 > 0 Dr. Karim Kamal 17 Graphs of Exponential Functions ❖ Which of these graphs is not corresponding to an exponential function? ❖ Note how the exponential growth is exceeding the polynomial of second degree. Dr. Karim Kamal 18 Properties of the Natural Exponential Function ❖ For all numbers 𝑥, 𝑥1 and 𝑥2 , the natural exponential 𝑒 𝑥 obeys the following laws: 1. 𝑒 𝑥1 · 𝑒 𝑥2 = 𝑒 𝑥1 +𝑥2 → 𝑒 2 𝑒 3 = 𝑒 5 2. 𝑒 −𝑥 = 1Τ𝑒 𝑥 → 𝑒 −3 = 1Τ𝑒 3 3. 𝑒 𝑥1 Τ𝑒 𝑥2 = 𝑒 𝑥1 −𝑥2 → 𝑒 5 Τ𝑒 3 = 𝑒 2 4. 𝑒 𝑥1 𝑥2 = 𝑒 𝑥1 𝑥2 → 𝑒 2 3 = 𝑒6 ❖ Note that these laws are not only valid for natural exponential functions but also for all exponential functions. Dr. Karim Kamal 19 The Logarithmic Function ❖ For any positive real number 𝑎 ≠ 1, and 𝑥 > 0, log 𝑎 𝑥 is the inverse of 𝑎 𝑥 ❖ Hence, we say that 𝑦 = log 𝑎 𝑥 iff 𝑥 = 𝑎 𝑦 ❖ Example: If 𝑦 = log 2 32, then 32 = 2𝑦 , which implies that 𝑦 = 5 ❖ So, we may look at log 𝑎 𝑥 as the power of 𝑎 that gives 𝑥 ❖ Examples: ❖ log 2 32 = 5 → because 25 = 32 ❖ log 2 2 = 1 → because 21 = 2 ❖ log 2 1 = 0 → because 20 = 1 Dr. Karim Kamal 20 Graph of the Logarithmic Function ❖ Note that the reflection of 𝑦 = 2𝑥 on the line 𝑦 = 𝑥 is indeed 𝑦 = log 2 𝑥 ❖ The line 𝑦 = 𝑥 is acting like a mirror for the two graphs 𝑦 = 2𝑥 and 𝑦 = log 2 𝑥 Dr. Karim Kamal 21 Graph of the Logarithmic Function ❖ The domain of all logarithmic functions log 𝑏 𝑎 𝑖𝑠 (𝑎 > 0) ❖ log 𝑏 1 = 0 for any positive base 𝑏 Dr. Karim Kamal 22 The Logarithmic Function: Special Cases ❖ The natural logarithmic function is the logarithmic function when its base is the number 𝑒 It is denoted by 𝑓 𝑥 = log 𝑒 𝑥 = ln 𝑥 ❖ The common logarithmic function is the logarithmic function when its base is the number 10 It is denoted by 𝑓 (𝑥) = log10 𝑥 = log 𝑥 Dr. Karim Kamal 23 The Graph of 𝑦 = ln 𝑥 Dr. Karim Kamal 24 Inverse Equations for 𝑎 and log 𝑎 𝒙 𝑥 ❖ Since 𝑎 𝑥 is the inverse of log 𝑎 𝑥, then we have the following results: 𝑎log𝑎 𝑥 = 𝑥 , (𝑥 > 0) → 3log3 5 = 5 log 𝑎 (𝑎 𝑥 ) = 𝑥, (𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥) → log 3 35 = 5 ❖ Special cases include: 𝑒 ln 𝑥 = 𝑥 , (𝑥 > 0) → 𝑒 ln 5 = 5 ln 𝑒 𝑥 = 𝑥, (𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥) → ln 𝑒 5 = 5 Dr. Karim Kamal 25 Rules of Logarithms ❖ 1. Product rule log 𝑎 𝑥𝑦 = log 𝑎 𝑥 + log 𝑎 𝑦 , 𝑎𝑙𝑠𝑜 ln 𝑥𝑦 = ln 𝑥 + ln 𝑦 ❖ 2. Quotient rule 𝑥 𝑥