Questions and Answers #1 Algebra I Functions Problems

Questions and Answers Sheet 1 Vectors and Spaces Question #1 1 3 Determine whether 𝑓(𝑥) = √5𝑥 + 10 and 𝑔(𝑥) = 𝑥 3 − 2 are inverse functions. 5 The functions are inverse The functions are not inverses. Answer: Step 1 3 1 The given functions are 𝑓(𝑥) = √5𝑥 + 10 and 𝑔(𝑥) = 𝑥 3 − 2 5 To check whether these functions are inverse functions, one can either start with f(x) or g(x) find their respective inverse and show that is equal to another function. 1 So, lets start with 𝑔(𝑥) = 𝑥 3 − 2 5 Write y=g(x) 1 𝑦 = 𝑥3 − 2 5 Step 2 Interchange x and y in the equation 1 𝑥 = 𝑦3 − 2 5 Now, find y in terms of x. 1 𝑥 = 𝑦3 − 2 5 1 𝑥 + 2 = 𝑦3 5 5𝑥 + 10 = 𝑦 3 3 𝑦 = √5𝑥 + 10 3 The y obtained is equal to the 𝑓(𝑥) = √5𝑥 + 10 So, f and g are inverse functions. Question #2 In each of Problems 1 through 5, find the Wroian of the given pair of functions. 1. 𝑒 2𝑡 , 𝑒 −3𝑡 2 Answer: Step 1 To determine Find the Wroian of the given pair of functions. 𝑒 2𝑡 , 𝑒 −3𝑡 2 Step 2 Formula used Let f(x) & g(x) be two functions. Then, the wroian of the two functions is given by: W ( f , g) = Step 3 f ( x) g ( x) f ( x) g ( x) Calculation −3𝑡 𝑓(𝑡) = 𝑒 2𝑡 , 𝑔(𝑡) = 𝑒 2 The wroina is given by: e W ( f , g) = 2t 2e e −3t 2 −3 −23t e 2 2t −3𝑡 −3 −3𝑡 ⇒ 𝑊(𝑓, 𝑔) = 𝑒 2𝑡 ( 𝑒 2 ) − (𝑒 2 ) (2𝑒 2𝑡 ) 2 ⇒ 𝑊(𝑓, 𝑔) = ⇒ 𝑊(𝑓, 𝑔) = −3 2 −3 2 −3𝑡 2 𝑒 2𝑡− 3𝑡 − 2𝑒 2𝑡− 2 𝑡 𝑡 𝑒 2 − 2𝑒 2 𝑡 ⇒ 𝑊(𝑓, 𝑔) = 𝑒 2 ( −3 − 2) 2 𝑡 2 7 ⇒ 𝑊(𝑓, 𝑔) = 𝑒 (− ) 2 7 𝑡 Hence, wroian of the given pair of functions is − 𝑒 2 2 Question #3 9 The formula 𝑦 = 𝑓(𝑥) = ( ) 𝑥 + 32 is used to convert from x degrees Celsius to y degrees Fahrenheit. 5 5 The formula 𝑦 = 𝑔(𝑥) = (𝑥 − 32) is used to convert from x degrees Fahrenheit to y degrees Celsius. 9 Show that f and g are inverse functions. Answer: Step 1 To determine Given: 9 5 5 9 𝑓(𝑥) = ( ) 𝑥 + 32 & 𝑔(𝑥) = (𝑥 − 32) To show: f & g are inverse functions. Step 2 Proof To show that f & g are inverse functions, we will show that (𝑓 ⋅ 𝑔)(𝑥) = (𝑔 ⋅ 𝑓)(𝑥) = 𝑥 Consider (𝑓 ⋅ 𝑔)(𝑥) =f(g(x)) 5 = 𝑓 ( (𝑥 − 32)) 9 9 5 5 9 = ( ) ( (𝑥 − 32)) + 32 =x-32+32 =x So, (𝑓 ⋅ 𝑔)(𝑥) = 𝑥 Consider (𝑔 ⋅ 𝑓)(𝑥) =g(f(x)) 9 = 𝑔 (( 𝑥) + 32) 5 5 9 = (( 𝑥) + 32 − 32) 9 5 5 9 = ( 𝑥) 9 5 =x So, (𝑔 ⋅ 𝑓)(𝑥) = 𝑥 Since, (𝑓 ⋅ 𝑔)(𝑥) = (𝑔 ⋅ 𝑓)(𝑥) therefore, f & g are inverse functions of each other. Question #4 Select the correct graph, showing f and g are inverse functions. 𝑥−1 𝑓(𝑥) = 6𝑥 + 1, 𝑔(𝑥) = 6 Answer: Step 1 Here, we need to show that 𝑓(𝑥) = 6𝑥 + 1, 𝑔(𝑥) = 𝑥−1 6 are inverse functions. Now, 𝑓(𝑔(𝑥)) = 𝑓 ( = 6( 𝑥−1 6 𝑥−1 6 ) )+1 =x Step 2 and 𝑔(𝑓(𝑥)) = 𝑔(6𝑥 + 1) = 6𝑥+1−1 6 =x Thus, 𝑓(𝑔(𝑥)) = 𝑔(𝑓(𝑥)) = 𝑥 Answer: 𝑥−1 Hence 𝑓(𝑥) = 6𝑥 + 1, 𝑔(𝑥) = are inverse functions. 6 Question #5 3 Write the function ℎ(𝑥) = √8−𝑥 5 as a composition of two functions Answer: Step 1 3 A function whose input is another function is a composite function. The function ℎ(𝑥) = √8−𝑥 5 if to be written as a composite function has to be separated into two functions. Let the first function be g(x) and other function be p(x). Then p(g(x))=h(x). Step 2 3 Let g(x)=8-x , Let 𝑝(𝑥) = √𝑥 5 Find p(g(x)) 𝑝(𝑔(𝑥)) = 𝑝(8 − 𝑥) 3 = √8−𝑥 5 =h(x) 3 Therefore h(x) is written as a composition of two functions g(x)=8-x and 𝑝(𝑥) = √𝑥 5 Question #6 Derivatives of logarithmic functions Calculate the derivative of the following functions. In some cases, it is useful to use the properties of logarithms to simplify the functions before computing f'(x). 𝑦 = log 2 (log 2 𝑥) Answer: Given, 𝑦 = log 2 (log 2 𝑥) On simplification, we get ln(log 2 𝑥) 𝑦= ln 2 ln 𝑥 [∵ log 𝑏 𝑥 = ] ln 𝑏 Now differentiating with respect to x, we get 𝑑𝑦 𝑑 ln(log 2 𝑥) = ( ) 𝑑𝑥 𝑑𝑥 ln 2 1 𝑑(ln(log 2 𝑥)) = ⋅ ln 2 𝑑𝑥 1 1 𝑑 ln 𝑥 𝑑(ln 𝑥) 1 = ⋅ ⋅ ( ) [∵ = ] = = = ln 2 1 ln 2 1 ln 2 1 ln 2 ⋅ ⋅ ⋅ log2 𝑥 1 log2 𝑥 1 log2 𝑥 1 log2 𝑥 ⋅ ⋅ ⋅ 𝑑𝑥 ln 2 𝑑 ln 𝑥 ( 𝑑𝑥 ) [∵ log 𝑏 𝑥 = 𝑑𝑥 ln 2 1 𝑑(ln 𝑥) ln 2 1 ln 2 ⋅ ⋅ 𝑥 ln 𝑥 ln 𝑏 ] 𝑑𝑥 1 𝑥 1 1 = ⋅ 2 𝑥(ln 2) log 2 𝑥 1 1 = ⋅ 𝑥(ln 2)2 ln 𝑥 ln 2 1 = 𝑥(ln 𝑥)(ln 2) Question #7 Derivatives of logarithmic functions Calculate the derivative of the following functions. In some cases, it is useful to use the properties of logarithms to simplify the functions before computing f'(x). 𝑦 = log 8 |tan 𝑥| Answer: Step 1 Derivative: 𝑦 = log 8 |tan 𝑥| log 𝑎 𝑏 = ln 𝑏 ln 𝑎 𝑦 = log 8 |tan 𝑥| = ln(tan 𝑥) ln 8 Step 2 𝑑𝑦 𝑑𝑥 = = = = 1 𝑑 ln 8 𝑑𝑥 1 sec2 𝑥 (ln(tan 𝑥)) ln 8 tan 𝑥 1 sec2 𝑥 ln 23 tan 𝑥 1 sec2 𝑥 3 ln 2 tan 𝑥 Question #8 Derivatives of logarithmic functions Calculate the derivative of the following functions. In some cases, it is useful to use the properties of logarithms to simplify the functions before computing f'(x). 𝑓(𝑥) = ln(3𝑥 + 1)4 Answer: Step 1 We first use the exponential rule of log. 𝑓(𝑥) = ln(3𝑥 + 1)4 𝑓(𝑥) = 4 ln(3𝑥 + 1) Step 2 Then we differentiate using the chain rule 𝑓(𝑥) = 4 ln(3𝑥 + 1) 1 𝑑 (3𝑥 + 1) 𝑓 ′ (𝑥) = 4 (3𝑥+1) 𝑑𝑥 1 𝑓 ′ (𝑥) = 4 (3𝑥+1) (3) 𝑓 ′ (𝑥) = (3𝑥+1) 12 12 Answer: 𝑓 ′ (𝑥) = (3𝑥+1)