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Question 10(Multiple Choice Worth 10 points)
(03.03 MC)

Select which function f has an inverse g that satisfies g prime of 2 equals 1 over 6 period

f(x) = 2x3
f of x equals 1 over 8 times x cubed
f(x) = x3
1 over 3 times x cubed

Sagot :

xero099

[tex]f(x) = 2\cdot x^{3}[/tex] has an inverse [tex]g[/tex] that satisfies [tex]g'(2) = \frac{1}{6}[/tex].

Procedure - Inverse of a function

First, we determine the value of [tex]x[/tex] for [tex]f(x)[/tex] in each case, based on the fact that [tex]f(x) = 2[/tex]:

Case A

[tex]2\cdot x^{3} = 2[/tex]

[tex]x = 1[/tex]

Case B

[tex]2 = \frac{1}{8}\cdot x^{3}[/tex]

[tex]16=x^{3}[/tex]

[tex]x = \sqrt[3]{16}[/tex]

Case C

[tex]2 = x^{3}[/tex]

[tex]x = \sqrt[3]{2}[/tex]

Case D

[tex]2 = \frac{1}{3}\cdot x^{3}[/tex]

[tex]6 = x^{3}[/tex]

[tex]x = \sqrt[3]{6}[/tex]

Now, we proceed to switch variables and find the derivative of the inverse function, whose formula is described below:

[tex]g'(x) = \frac{1}{f'(y)}[/tex] (1)

Where:

  • [tex]y[/tex] - Dependent variable of the inverse function (Independent variable of the original function).
  • [tex]x[/tex] - Independent variable of the inverse function (Dependent variable of the original function).

Case A ([tex]x = 2[/tex], [tex]y = 1[/tex])

[tex]f(y) = 2\cdot y^{3}[/tex]

[tex]f'(y) = 6\cdot y^{2}[/tex]

[tex]f'(1) = 6\cdot 1^{2}[/tex]

[tex]f'(1) = 6[/tex]

[tex]g'(2) = \frac{1}{6}[/tex] [tex]\blacksquare[/tex]

Case B ([tex]x = 2[/tex], [tex]y = \sqrt[3]{16}[/tex])

[tex]f(y) = \frac{1}{8}\cdot y^{3}[/tex]

[tex]f'(y) = \frac{3}{8}\cdot y^{2}[/tex]

[tex]f'(\sqrt[3]{16}) = \frac{3}{8}\cdot \sqrt[3]{4}[/tex]

[tex]g'(2) = \frac{8}{3\cdot \sqrt[3]{4}}[/tex]

[tex]g'(2) = \frac{8\cdot \sqrt[3]{16}}{12}[/tex] [tex]\blacksquare[/tex]

Case C ([tex]x = 2[/tex], [tex]y = \sqrt[3]{2}[/tex])

[tex]f(y) = y^{3}[/tex]

[tex]f'(y) = 3\cdot y^{2}[/tex]

[tex]f'(\sqrt[3]{2}) = 3\cdot \sqrt [3]{4}[/tex]

[tex]g'(2) = \frac{1}{3\cdot \sqrt [3]{4}}[/tex]

[tex]g'(2) = \frac{\sqrt [3]{16}}{12}[/tex] [tex]\blacksquare[/tex]

Case D ([tex]x = 2[/tex], [tex]y = \sqrt[3]{6}[/tex])

[tex]f(y) = \frac{1}{3}\cdot y^{3}[/tex]

[tex]f'(y) = y^{2}[/tex]

[tex]f(\sqrt[3]{6}) = \sqrt[3]{36}[/tex]

[tex]g'(2) = \frac{1}{\sqrt [3]{36}}[/tex]

[tex]g'(2) = \frac{36^{2/3}}{36}[/tex]

[tex]g'(2) = \frac{\sqrt[3]{1296}}{36}[/tex] [tex]\blacksquare[/tex]

In a nutshell, [tex]f(x) = 2\cdot x^{3}[/tex] has an inverse [tex]g[/tex] that satisfies [tex]g'(2) = \frac{1}{6}[/tex]. [tex]\blacksquare[/tex]

To learn more on inverse functions, we kindly invite to check this verified question: https://brainly.com/question/5245372

Remarks

The statement is poorly formated, correct form is presented below:

Select which function [tex]f[/tex] has an inverse [tex]g[/tex] that satisfies [tex]g'(2) = \frac{1}{6}[/tex]:

A. [tex]f(x) = 2\cdot x^{3}[/tex]

B. [tex]f(x) = \frac{1}{8}\cdot x^{3}[/tex]

C. [tex]f(x) = x^{3}[/tex]

D. [tex]f(x) = \frac{1}{3}\cdot x^{3}[/tex]

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