Mathematics, Science and Technology Midterm Exam PDF

MATHEMATICS, SCIENCE AND TECHNOLOGY Prepared by: Jolina A. Ramos, LPT, MAED units INVERSE FUNCTIONS Objective: Define inverse function. Find the inverse of a given function. Graph the given function and its inverse. INVERSE FUNCTIONS Is a function 𝑓 −1 such that 𝑓[𝑓 −1 𝑥 ] for each value...

MATHEMATICS, SCIENCE AND TECHNOLOGY Prepared by: Jolina A. Ramos, LPT, MAED units INVERSE FUNCTIONS Objective: Define inverse function. Find the inverse of a given function. Graph the given function and its inverse. INVERSE FUNCTIONS Is a function 𝑓 −1 such that 𝑓[𝑓 −1 𝑥 ] for each value of x in the domain of 𝑓 −1 and 𝑓 −1 𝑓 𝑥 = 𝑥 for each value of x in the domain of f. Thus, if 𝑓 −1 is the inverse f, then we have 𝑓 −1 = 𝑓 and 𝑓 = 𝑓 −1. Find the inverse of f = 𝟏, 𝟓 , 𝟐, 𝟔 , 𝟑, 𝟕 , ( 𝟒, 𝟖) −𝟏 𝒇 = 𝟖, −𝟏 , 𝟎, 𝟕 , 𝟕, 𝟐 , ( 𝟖, 𝟓)  EXAMPLES! Find the inverse of the following equations. 𝟏. 𝒇 𝒙 = 𝟐𝒙 − 𝟓 3. 𝒇 𝒙 = 𝟑𝒙 + 𝟐 2. 𝒇 𝒙 = 𝟑𝒙 + 𝟔 4. 𝒇 𝒙 = 𝟐𝒙 − 𝟒 𝟐  X 0 -1 -2 -3 f(x) = 3x + 6 y 6 3 0 -3 𝒇 𝒙 = 𝟑𝒙 + 𝟔 𝒇 𝒙 = 𝟑𝒙 + 𝟔 𝒇 𝒙 = 𝟑𝒙 + 𝟔 𝒇 𝒙 = 𝟑 −𝟐 + 𝟔 𝒇 𝒙 =𝟑 𝟎 +𝟔 𝒇 𝒙 = 𝟑 −𝟏 + 𝟔 𝒇 𝒙 = −𝟑 + 𝟔 𝒇 𝒙 = −𝟔 + 𝟔 𝒇 𝒙 =𝟔 𝒇 𝒙 =𝟎 𝒇 𝒙 =𝟑 𝒇 𝒙 = 𝟑𝒙 + 𝟔 𝒇 𝒙 = 𝟑 −𝟑 + 𝟔 The ordered pairs of (x, y) = 𝒇 𝒙 = −𝟗 + 𝟔 (0,6) , (-2,0) , (-1, 3), (-3, -3) −1 𝒇 𝒙 = −𝟑 The ordered pairs of 𝑓 =(6,0) , (0,-2) , (3,-1), (-3, -3)  −𝟏 𝒙−𝟔 𝒇 (x) = 𝟑 f(x) = 3x + 6 EXPONENTIAL FUNCTIONS LAWS OF EXPONENT MULTIPLICATION RULE ax ay = ax+y - if the operation is Example: m7 m3 multiplication, add the exponents. Solution: m7+3 = m10 DIVISION RULE ax = ax−y - if the operation is Example: a10 ay a6 division, subtract the exponents Solution: a10−6 =a 4 POWER OF A POWER RULE ax y = axy - in this rule, just simply Example: a7 2 multiply exponent to exponent Solution: a7 2 = a14 POWER OF A PRODUCT RULE ab x = 𝑎 𝑥 𝑏 𝑥 – distribute the Example: ab 4 exponents on the base Solution: a4 b4 POWER OF A FRACTION RULE a x ax a 3 b = bx - the same with the product Example: b rule of distribution of exponents a3 Solution: b3 ZERO EXPONENT a0 = 1 - all number or base with zero Example: 120 exponents is equals to 1 Solution:1 NEGATIVE EXPONENT −x 1 a = - in order to eliminate negative exponent, get the ax reciprocal, it should be in the fractional form which is 1 is a constant as numerator and the base with negative exponents is the denominator. Example: a−7 1 Solution: 7 a  FRACTIONAL EXPONENT x 3 y a = ax - if the exponent is fraction, y Example: 25 𝟓 write it on the radical form which is the Solution: 𝟐𝟑 numerator will be the exponent and the 𝟓 𝟖 denominator will be the index. y Index ax exponent EVALUATION EXPONENTIAL FUNCTIONS An exponential function with base b is a function of the form f(x) =𝑏^𝑥 or y=𝑏^𝑥 (b > 0, b ≠ 1). To evaluate exponential functions, substitute the value of the variable in the function 𝑓 𝑥 = 𝑎 𝑥. EXAMPLES 1 2−𝑥 3 If f(x)=2 , 𝑔 𝑥 = 𝑥 and ℎ 𝑥 = 3−𝑥−1 find f 4 2 Solution: 𝟑 𝟑 = 𝟒∙𝟐 𝒇 = 𝟐𝟐 𝟐 =𝟐 𝟐 = 𝟐𝟑 = 𝟖 EXAMPLES 1 2−𝑥 If f(x)=2𝑥 , 𝑔 𝑥 = and ℎ 𝑥 = 3−𝑥−1 find f −2 4 Solution: −𝟐 𝟐−𝒙 −𝟏 𝒇 −𝟐 = 𝟐 𝟏 𝟏 𝒈 𝟑 = 𝟒 𝟒 𝟏 = 𝟐 𝟐−𝟑 𝟐 𝟏 𝟒 𝟏 𝟒 = 𝟒 EXPONENTIAL EQUATIONS EXPONENTIAL EQUATIONS Are equations whose exponents are variables. To solve an exponential equation, express both sides of the equation as powers with the same base, then solve the resulting equation. EXPONENTIAL EQUATIONS 𝑥 𝑥 2 =8 1 1 2𝑥 = 2 = 64 16 2𝑥 = 23 𝑥 2 =2 −4 2−1(𝑥) = 64 𝑥=3 𝑥 = −4 2−1(𝑥) = 26 −𝑥 = 6 𝑥 = −6 EXPONENTIAL EQUATIONS 2𝑥 = 128 32𝑥 = 243 2𝑥−2 = 64 𝑥 7 32𝑥 = 35 2𝑥−2 = 26 2 =2 2𝑥 = 5 𝑥−2=6 𝑥=7 5 𝑥 =6+2 𝑥=8 𝑥= 2 APPLICATION OF EXPONENTIAL FUNCTIONS  Exponential Growth Exponential growth is a process that increases quantity over time at an ever-increasing rate. Exponential Decay Describes the process of reducing an amount by a consistent percentage rate over a period of time Growth Decay A colony of bacteria in a human body The concentration of medicine that a patient took as time passed by. Population in the Philippines Compound Interest Value of Gadgets Over Year Remains of a Radioactive Element  Exponential Growth An exponential function of the form 𝑓 𝑥 = 𝑘𝑎 𝑥 , 𝑤ℎ𝑒𝑟𝑒 𝑎 > 1 𝑎𝑛𝑑 𝑘 > 0, the value of f(x) increases without bound. Exponential Decay An exponential function of the form 𝑓 𝑥 = 𝑘𝑎 𝑥 , 𝑤ℎ𝑒𝑟𝑒 𝑎 0 < 𝑎 < 1 𝑎𝑛𝑑 𝑘 > 0, the value of f(x) decreases. Which of the following functions exhibits exponential decay? a. 𝑓 𝑥 = 3𝑥 Exponential Growth b. 𝑔 𝑥 = 𝟏 𝒙 Exponential Decay 𝟒 c. h 𝑥 = 2−𝑥 Exponential Decay 𝟓 𝒙 d. i x = 𝟐 Exponential Growth LOGARITHMIC FUNCTIONS  Logarithmic Functions The logarithmic function is the function 𝑦 = 𝑙𝑜𝑔𝑏 𝑥 where b is a real number such that 𝑏 > 0, 𝑏 ≠ 1 and 𝑦 𝑥 > 0. And 𝑦 = 𝑙𝑜𝑔𝑏 𝑥, if and only if, 𝑥 = 𝑏. 𝑙𝑜𝑔3 𝑥 = 4 𝑙𝑜𝑔3 81 = 𝑥 𝑙𝑜𝑔10 0.01 = −2 𝑙𝑜𝑔𝑒 9 = 𝑦 𝑙𝑜𝑔5 𝑥 = 4  Writing Logarithmic Equation into Exponential Equations and Vice Versa 4 𝑙𝑜𝑔3 𝑥 = 4 3 =𝑥 𝑙𝑜𝑔3 81 = 𝑥 𝑥 3 = 81 𝑙𝑜𝑔10 0.01 = −2 10−2 = 0.01 𝑙𝑜𝑔𝑒 9 = 𝑦 𝑒 =9𝑦 Writing Exponential Form into Logarithmic Form 54 =𝑥 𝑙𝑜𝑔5 𝑥 = 4 6 2 = 64 𝑙𝑜𝑔2 64 = 6 𝑎7 = 𝑀 𝑙𝑜𝑔𝑎 𝑀 = 7 𝑥 𝑐 =𝑇 𝑙𝑜𝑔𝐶 𝑇 = 𝑥  Solving Logarithmic Equations 𝑙𝑜𝑔3 𝑥 = 2 𝑙𝑜𝑔3 81 = 𝑥 𝑙𝑜𝑔5 𝑥 = 4 2 3 =𝑥 𝑥 3 = 81 54 = 𝑥 9=𝑥 𝑥=4 625 = 𝑥 𝑙𝑜𝑔3 9 = 2 𝑙𝑜𝑔3 81 = 4 𝑙𝑜𝑔5 625 = 4  Try This! 1. 𝑙𝑜𝑔4 64 = 𝑥 4. 𝑙𝑜𝑔3 9 = 𝑥 2. 𝑙𝑜𝑔3 81 = 𝑥 5. 𝑙𝑜𝑔10 0.1 = 𝑥 3. 𝑙𝑜𝑔1 8 = 𝑥 6. 𝑙𝑜𝑔1 16 = 𝑥 2 2 LAWS OF LOGARITHM  LAWS OF LOGARITHM MULTIPLICATION RULE FOR LOGARITHMS DIVISION RULE FOR LOGARITHMS POWER RULE FOR LOGARITHMS  MULTIPLICATION RULE FOR LOGARITHMS 𝒍𝒐𝒈𝒃 (𝑴𝑵) 𝒍𝒐𝒈𝒃 𝑴 + 𝒍𝒐𝒈𝒃 (𝑵) 𝒍𝒐𝒈𝟒 (𝑿𝒀) 𝒍𝒐𝒈𝟒 𝑿 + 𝒍𝒐𝒈𝟒 (𝒀) 𝒍𝒐𝒈𝟓 (𝟒 · 𝟓) 𝒍𝒐𝒈𝟓 𝟒 + 𝒍𝒐𝒈𝟓 (𝟓) 𝒍𝒐𝒈𝟑 (𝟐𝟓) 𝒍𝒐𝒈𝟑 𝟓 + 𝒍𝒐𝒈𝟑 (𝟓)  MULTIPLICATION RULE FOR LOGARITHMS 𝒍𝒐𝒈𝟑 𝑨 + 𝒍𝒐𝒈𝟑 (𝑩) 𝒍𝒐𝒈𝟑 (𝑨𝑩) 𝒍𝒐𝒈𝟐 𝟔 + 𝒍𝒐𝒈𝟐 (𝟖) 𝒍𝒐𝒈𝟐 (𝟒𝟖) 𝒍𝒐𝒈𝟓 𝟐 + 𝒍𝒐𝒈𝟓 (𝟑) 𝒍𝒐𝒈𝟓 (𝟔) 𝒍𝒐𝒈𝟔 𝑿 + 𝒍𝒐𝒈𝟔 (𝒀) 𝒍𝒐𝒈𝟔 (𝑿𝒀)  DIVISION RULE FOR LOGARITHMS 𝑿 𝒍𝒐𝒈𝒃 𝒍𝒐𝒈𝒃 𝑿 − 𝒍𝒐𝒈𝒃 (𝒀) 𝒀 𝟓 𝒍𝒐𝒈𝟑 𝒍𝒐𝒈𝟑 𝟓 − 𝒍𝒐𝒈𝟑 (𝟏𝟐) 𝟏𝟐 𝟏𝟎 𝒍𝒐𝒈𝟒 𝒍𝒐𝒈𝟒 𝟏𝟎 − 𝒍𝒐𝒈𝟒 (𝟐𝟎) 𝟐𝟎  DIVISION RULE FOR LOGARITHMS 𝑨 𝒍𝒐𝒈𝟓 𝑨 − 𝒍𝒐𝒈𝟓 (𝑩) 𝒍𝒐𝒈𝟓 𝑩 𝒍𝒐𝒈𝟓 𝟏𝟓 − 𝒍𝒐𝒈𝟓 (𝟓) 𝒍𝒐𝒈𝟓 𝟑 𝟑 𝒍𝒐𝒈𝟐 𝟑 − 𝒍𝒐𝒈𝟐 (𝟓) 𝒍𝒐𝒈𝟐 𝟓  POWER RULE FOR LOGARITHMS 𝒍𝒐𝒈𝟓 𝑴𝑵 𝑵 𝒍𝒐𝒈𝟓 𝑴 𝟏𝟎 𝟏𝟎 𝒍𝒐𝒈𝟒 𝟓 𝒍𝒐𝒈𝟒 (𝟓 ) 𝟏 𝟏 𝒍𝒐𝒈𝟐 𝟔𝟑 𝒍𝒐𝒈𝟐 𝟔 𝟑  POWER RULE FOR LOGARITHMS 𝑿 𝑿 𝒍𝒐𝒈𝑩 𝒀 𝒍𝒐𝒈𝑩 (𝒀 ) 𝟐 𝟐 𝒍𝒐𝒈𝟓 𝒀 𝒍𝒐𝒈𝟓 (𝒀 ) 𝟏 𝒍𝒐𝒈𝟓 𝟐 𝒍𝒐𝒈𝟓 ( 𝟐) 𝟐  TRY THIS: EXPRESS THE FOLLOWING INTO EXPANDED LOGARITHM 𝒍𝒐𝒈𝟒 𝒙 𝒚𝟐 𝒍𝒐𝒈𝒃 𝒃𝟐 𝒄𝒅𝟑 𝟐 𝒍𝒐𝒈𝒃 𝒃𝟐 + 𝒍𝒐𝒈𝒃 𝐜 + 𝒍𝒐𝒈𝒃 (𝒅𝟑 ) 𝒍𝒐𝒈𝟒 𝒙 + 𝒍𝒐𝒈𝟒 (𝐲) 𝟐 𝒍𝒐𝒈𝟒 𝒙 + 𝒍𝒐𝒈𝟒 (𝐲) 𝟐𝒍𝒐𝒈𝒃 𝒃 + 𝒍𝒐𝒈𝒃 𝐜 + 𝟑𝒍𝒐𝒈𝒃 (𝒅) TRY THIS: EXPRESS THE FOLLOWING INTO EXPANDED LOGARITHM 𝒙𝟐 𝒚𝟐 𝒍𝒐𝒈𝟓 𝒛𝟑 𝟐 𝟐 𝟑 𝒍𝒐𝒈𝟓 𝒙 + 𝒍𝒐𝒈𝟓 𝒚 − 𝒍𝒐𝒈𝟓 𝒛 𝟐 𝒍𝒐𝒈𝟓 𝒙 + 𝟐 𝒍𝒐𝒈𝟓 𝐲 − 𝟑𝒍𝒐𝒈𝟓 𝒛 TRY THIS: EXPRESS THE FOLLOWING INTO EXPANDED LOGARITHM 𝒍𝒐𝒈𝟓 𝒂𝟐 𝒃𝟑 𝒄𝟒 𝟐 𝟑 𝟒 𝒍𝒐𝒈𝟓 𝒂 + 𝒍𝒐𝒈𝟓 𝒃 + 𝒍𝒐𝒈𝟓 𝒄 𝟐 𝒍𝒐𝒈𝟓 𝒂 + 𝟑 𝒍𝒐𝒈𝟓 𝐛 + 𝟒𝒍𝒐𝒈𝟓 𝒄  TRY THIS: EXPRESS THE FOLLOWING INTO EXPANDED LOGARITHM 𝒄𝟑 𝒅𝟑 𝒍𝒐𝒈𝟓 𝟓 𝟓 𝒆 𝒇 𝟑 𝟑 𝟓 𝟓 𝒍𝒐𝒈𝟓 𝒄 + 𝒍𝒐𝒈𝟓 𝒅 − 𝒍𝒐𝒈𝟓 𝒆 +𝒍𝒐𝒈𝟓 𝒇 𝟑𝒍𝒐𝒈𝟓 𝒄 + 𝟑𝒍𝒐𝒈𝟓 𝐝 − 𝟓𝒍𝒐𝒈𝟓 𝒆 +𝟓𝒍𝒐𝒈𝟓 𝒇 TRY THIS: EXPRESS THE FOLLOWING INTO EXPANDED LOGARITHM 𝒍𝒐𝒈𝟓 𝒙𝒚 𝟏 𝒍𝒐𝒈𝟓 (𝒙𝒚)𝟐 𝟏 𝒍𝒐𝒈𝟓 (𝒙𝒚) 𝟐 𝟏 𝒍𝒐𝒈𝟓 𝒙 + 𝒍𝒐𝒈𝟓 𝒚 𝟐 TRY THIS: EXPRESS THE FOLLOWING INTO SINGLE LOGARITHM 𝒍𝒐𝒈𝟐 𝒙 + 𝒍𝒐𝒈𝟐 𝒚 − 𝟑𝒍𝒐𝒈𝟐 (𝒛) (𝒙𝒚) 𝒍𝒐𝒈𝟐 𝟑 𝒛 TRY THIS: EXPRESS THE FOLLOWING INTO SINGLE LOGARITHM 𝟑𝒍𝒐𝒈𝟒 𝒘 + 𝟐𝒍𝒐𝒈𝟒 𝒙 − 𝟑𝒍𝒐𝒈𝟒 𝒚 + 𝟒𝒍𝒐𝒈𝟒 (𝒛) 𝟑 𝟐 (𝒘 𝒙 ) 𝒍𝒐𝒈𝟒 𝟑 𝟒 (𝒚 𝒛 ) MATHEMATICS, SCIENCE AND TECHNOLOGY Prepared by: Jolina A. 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