General Mathematics PDF

GENERAL MATHEMATICS FIRST HALF-SEMESTER Week 1: Function SUBTRACTION - subtracting the first term to the second term. CAUTION: DISTRIBUTE the negative sign (-) to each term of RELATION -...

GENERAL MATHEMATICS FIRST HALF-SEMESTER Week 1: Function SUBTRACTION - subtracting the first term to the second term. CAUTION: DISTRIBUTE the negative sign (-) to each term of RELATION - a rule that relates values from a set of values the second function and then proceed to combine like terms. (called the domain) to a second set of values (called the (f - g) (x) = f(x) - g(x) range). A relation is a set of ordered pairs (x,y). Example: 3. If f(x) = x - 2 and g(x) = x² - 4, solve (f - g) (x) FUNCTION - a relation where each element in the domain is.Solution: related to only one value in the range by some rule. A ➔ (f - g) (x) = f(x) - g(x) function is a set of ordered pairs (x,y) such that no two ➔ (f - g) (x) = (x - 2) - (x² - 4) ordered pairs have the same x-value but different y-values. ➔ (f - g) (x) = x - 2 - x² + 4 (Meaning, bawal magkapareho ng x ang isang function pero ➔ (f - g) (x) = - x² + x + 2 pwede magkapareho ng y) 4. If f(x) = 2x + 3 and g(x) = 4x + 1, solve (f - g) (x) VERTICAL LINE TEST - a graph that represents a function only Solution: if each vertical line intersects the graph at most once. ➔ (f - g) (x) = f(x) - g(x) ➔ (f - g) (x) = (2x + 3) - (4x + 1) EVALUATING A FUNCTION - substituting the value of x. ➔ (f - g) (x) = 2x + 3 - 4x - 1 Example: ➔ (f - g) (x) = - 2x + 2 If f(x) = 3x+4, find f(-7) ➔ f(-7) = 3(-7)+4 MULTIPLICATION - Put parentheses around them and ➔ f(-7) = -21+4 multiply each term from the first function to each term of the ➔ f(-7) = -17 second function. (f · g) (x) = f(x) · g(x) 3. If f(x) = 6x+3 and g(x) = 4x+2, solve (f x g) (x) OPERATION OF FUNCTION Solution: ADDITION - adding the two functions together and simplify ➔ (f · g) (x) = f(x) · g(x) by finding the like terms. (f + g) (x) = f(x) + g(x) ➔ (f · g) (x) = (6x+3)(4x+2)(foil method) Example: ➔ (f · g) (x) = 24x²+12x+12x+6 1. If f(x) = 3x + 2 and g(x) = 5x + 3, solve (f + g) (x) ➔ (f · g) (x) = 24x²+24x+6 Solution: ➔ (f + g) (x) = f(x) + g(x) 4. If f(x) =2x+3 and g(x) = 4x+1, solve (f x g) (x) ➔ (f + g) (x) = (3x + 2) + (5x + 3) Solution: ➔ (f + g) (x) = 8x + 5 ➔ (f · g) (x) = f(x) · g(x) 2. If f(x) = 2x+1 and g(x) = x² - 9, solve (f + g) (x) ➔ (f · g) (x) = (2x+3)(4x+1) Solution: ➔ (f · g) (x) = 8x²+2x+12x+3 ➔ (f + g) (x) = f(x) + g(x) ➔ (f · g) (x) = x²+14x+3 ➔ (f + g) (x) = (2x + 1) + (x² - 9) ➔ (f + g) (x) = x² + 2x - 8 GENERAL MATHEMATICS FIRST HALF-SEMESTER DIVISION - To find the quotient of two functions, put the first 𝑓 𝑓(𝑥) one over the second. ( ) (x) = Week 2: Rational Function 𝑔 𝑔(𝑥) Lesson 2.1 Rational Function Example: Rational Function - A rational function is a function of the form 𝑓(𝑥) 𝑓 𝑓(𝑥) 3. If f(x) = 4x - 5 and g(x) = 2x, solve ( ) (x) = 𝑝(𝑥) / q(x) where𝑝(𝑥) and 𝑞(𝑥) are polynomial function and 𝑞(𝑥) is 𝑔 𝑔(𝑥) not the zero function. The domain of 𝑓(𝑥) is the set of all values of x Solution: where 𝑞(𝑥) ≠0. 𝑓 𝑓(𝑥) ➔ ( ) (x) = 𝑔 𝑔(𝑥) To find the domain, set the denominator equal to zero and solve 𝑓 4𝑥−5 for x. ➔ ( )= 𝑔 2𝑥 (nothing to do here) 2+𝑥 f(x) = (find the domain of the rational function) 𝑓 𝑥 4. If f(x) = 3x - 7 and g(x) = 5x, solve ( ) 𝑔 Solution: 2+𝑥 Solution: f(x) = (set the denominator value into zero and 𝑓 𝑓(𝑥) 𝑥 ➔ ( ) (x) = solve for x) 𝑔 𝑔(𝑥) 𝑓 3𝑥−7 ➔ 𝑔 ) (x) = 5𝑥 (nothing to do here) x = 0 (set the set-builder notation) {x ε R│x ≠ 0} COMPOSITE FUNCTION - a composite function is created when one function is substituted into another function. ³ˣ² ⁻ ˣ ⁺ ⁴ Example: Example 1: ₓ₊₇ 3. If f(x) = x2+4 and g(x) = 3x3+1, solve (g x f) Asymptotes: Solution: VA HA ➔ (f ◦ g) (x) = x2+4 x+7=0 y = NONE ➔ (f ◦ g) (x) = (3x3+1)2 + 4 x = -7 ➔ (f ◦ g) (x) = (3x3+1)(3x3+1)+4 X - Intercept & Zeroes: ➔ (f ◦ g) (x) = 9x6+3x3+3x3+1+4 ➔ (f ◦ g) (x) = 9x6+6x3+5 3x² - x + 4 = 0 (x + 1)(3x - 4) = 0 4. If f(x) = 3x+2 and g(x) = x2- 5, solve 2g(x) - 3f(x) x + 1 = 0 3x - 4 = 0 Solution: 4 x = -1 x = 3 ➔ 2g(x) - 3f(x) = 2(x2- 5) - 3(3x+2) ➔ 2g(x) - 3f(x) = 2(x2- 10 - 9x - 6) ➔ 2g(x) - 3f(x) = 2x2- 10 - 9x - 6 Y - Intercepts: ➔ 2g(x) - 3f(x) = 2x2- 9x - 10 - 6(-10 - 6 = -10 + -6 -16) 4 y = ³ˣ² ⁻ ˣ ⁺ 𝑥 ₊ ₇ ➔ 2g(x) - 3f(x) = 2x2- 9x - 16 GENERAL MATHEMATICS FIRST HALF-SEMESTER 4 1. Eliminate the rational expressions in the equation by y = ³⁽⁰⁾ ⁻ ⁽⁰⁾ ⁺ 0 ₊ ₇ multiplying both sides of the equation by the LCD. 4 2. Solve the equation. y= 7 3. Check your solution. Rational Functions: 2 9 x+7=0 {x ε R│x ≠ -7} Example 1: 𝑥₊ ₆ = 2 (ANG GULO PS. GALING 𝑥 ₊ ₄ₓ ₋ ₁₂ SA IBANG SECTION - kale) 2 9 3 𝑥₊ ₆ X 2 (cross multiplication) Example 2: f(x) = ₋₈ 𝑥 ₊ ₄ₓ ₋ ₁₂ 2 𝑥 9(x + 6) = 2(ₓ² ₊ ₄ₓ ₋ ₁₂) Asymptotes: 9x + 54 = 2x² + 8x - 24 VA HA − 9𝑥 + 54 x² - 8 = 0 y=0 = 2𝑥² + 8𝑥 − 24 √x² = √8 (x + 6) (2x - 13) X - Intercept & Zeroes: 3=0 x+6=0 2x - 13 = 0 x = NONE 2𝑥 13 X = -6 2 = 2 Y - Intercepts: 3 13 y= 2 ₋₈ x= 2 𝑥 3 𝑥 1 3 Example 2: + 9 = 4 = 2 ₋₈ 6 (0) LCD = 36 3 = −8 3 𝑥 1 36 [ 6 + 9 = 4 ] 36 −3 y= 8 18 + 4x = 9 4x = 9 - 18 Rational Functions: 4x = -9 X² - 8 = 0 4𝑥 −9 √x² = √8 = 4 4 x = ⁺₋ 2 √2 −9 {x ε R│x ≠ ⁺₋ 2 √2 X= 4 3 5 Week 3: Rational Equation & Inequality Example 2: 𝑥+2 = 𝑥+4 Lesson 2.2 Rational Equation 3 5 Rational Equation is an equation that contains one or more 𝑥+2 X 𝑥+4 (Cross Multiply) rational expressions. Rules on how to solve Rational Equation: 5(x+2) = 3(x+4) 5x-3x = 12 - 10 GENERAL MATHEMATICS FIRST HALF-SEMESTER 2x = 2 1 −2 ≤ 0 or - 0.5 ≤ 0 ←TRUE [ ] 2𝑥 2 2 = 2 X=1 Let x be 6 3(6) + 6 2(6) − 12 ≤0 24 0 ≤ 0 ←UNDEFINED () Whole Answer = [-2, 6 ) Lesson 2.3 Rational Inequality Example 2: Rational Inequality - is an inequality that contains rational expressions. Week 4: Inverse Function 3𝑥 + 6 Example 1: 2𝑥 − 12 ≤ 0 (must always be zero) 1. Replace f(x) with y Solution: 2. Interchange x and y Numerator: Denominator: 3. Solve the equation for y 3x+6 = 0 2x - 12 = 0 4. Replace y with f-1(x) 3x = -6 2x = 12 3𝑥 −6 2𝑥 12 (bawal makalimutan na hindi ma-replace ‘ung y sa f-1(x) // cha) = = 3 3 2 2 X = -2 x=6 EXAMPLE 1: f(x) = 4𝑥+2 5 4𝑥+2 y= 5 ←—---(-2)----------------(0)----------------(6)------------> 4𝑦+2 x= 5 [ ] - Part of the Solution 5x=4y+2 ( ) - not part of the solution 5x-2=4y ← divide both sides by 4 to cancel 4 from 4y 5𝑥−2 y= 4 Let x be -2 5𝑥−2 f–1(x)= 4 3(−2) + 6 2(−2) − 12 ≤0 0 −16 ≤0 0≤0 ←TRUE [ ] Let x be 0 3(0) + 6 2(0) − 12 ≤0 6 −12 ≤0 GENERAL MATHEMATICS FIRST HALF-SEMESTER EXPONENTIAL INEQUALITIES 1. If the same number is added to or subtracted from both Week 5: Exponential Function sides of an inequality (not changed). 2. If both sides of an inequality are multiple or divided by EXPONENTIAL FUNCTION - a form of f(x) = b^x, where "x" is a the same positive real number (not changed). variable and "b" is a constant which is called the base of the 3. If both sides of an inequality are multiple or divided by function and it could be greater than 0 the same negative real number (changed). DOMAIN - set of all real numbers { ∞ , -∞ } 4. If it’s fraction (change na agad since its less than 1 //cha) RANGE - set of positive real numbers { 0 , -∞} steps: ZEROES- refers to the value of the independent variable x that 1. Same base makes the function 0. (same cla ni x-intercept) 2. Bring down the exponent 3. Solve for X X-intercept - y=0 Example1: Y-intercept - x=0 43x+2