Question

for the function y = f(x) = 2x³ - 10:
a. find $\frac{df}{dx}$ at x = 4.
f(4) = 96

b. find a formula for x = f⁻¹(y).
f⁻¹(y) =

c. find $\frac{df^{-1}}{dy}$ at y = f(4).
(f⁻¹)(f(4)) =

Explanation:

Step1: Differentiate the function

Given $y = f(x)=2x^{3}-10$. Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $f'(x)=\frac{d}{dx}(2x^{3}-10)=2\times3x^{2}=6x^{2}$. Then $f'(4)=6\times4^{2}=6\times16 = 96$.

Step2: Find the inverse function

Set $y = 2x^{3}-10$. First, solve for $x$:
\[

$$\begin{align*} y&=2x^{3}-10\\ y + 10&=2x^{3}\\ x^{3}&=\frac{y + 10}{2}\\ x&=f^{-1}(y)=\sqrt[3]{\frac{y + 10}{2}} \end{align*}$$

\]

Step3: Use the formula for the derivative of the inverse function

The formula for the derivative of the inverse function is $(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}$. We know that $y = f(4)=2\times4^{3}-10=2\times64 - 10=128 - 10 = 118$. Since $f'(x)=6x^{2}$, and $f^{-1}(y)=\sqrt[3]{\frac{y + 10}{2}}$, then $(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}$. When $y = f(4)=118$, $f^{-1}(118)=4$. So $(f^{-1})'(f(4))=\frac{1}{f'(4)}$. Since $f'(4)=96$, then $(f^{-1})'(f(4))=\frac{1}{96}$.

Answer:

a. $96$
b. $\sqrt[3]{\frac{y + 10}{2}}$
c. $\frac{1}{96}$