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Question

# If g is the inverse function of f and $f(x)=x^3+3 x-1$, what is the value of g'(35)?

Solution

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Given that $g$ is the inverse of $f$ and

$f(x) = x^3 + 3x - 1$

The function $f$ must be a one-to-one function because its inverse is a function, and only one-to-one functions have inverses that are functions. Since $g$ is the inverse of $f$, we have

\begin{align*} g(x) & = f^{-1} (x)\\ f(g(x)) & = x\\ \dfrac{d}{dx} f(g(x)) & = \dfrac{d}{dx}x\\ f'(g(x)) g'(x)& =1 \\ g'(x) & = \dfrac{1}{f'(g(x))} \end{align*}

Thus, we begin by finding the derivative of $f$

\begin{align*} f(x) & = x^3 + 3x - 1\\ f'(x) & = 3x^2 + 3 \end{align*}

Substituting in the equation for the derivative of $g$ gives us

\begin{align*} g'(x) & = \dfrac{1}{3(g(x))^2 + 3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{substituted}\\ g'(35) & = \dfrac{1}{3(g(35))^2 + 3}\ \ \ \ \ \ \ \ \ \ \ \ \ \text{evaluated}\\ \end{align*}

Since $g$ is the inverse function of $f,$ we can say that $g(35)$ is the value of $x$ for which $f(x) = 35$. Thus, we need to solve

$35 = x^3 + 3x - 1$

Usually we would have to solve this cubic for $x$. but the solution is apparent by inspection. We can see that $x$ has to equal $3$.

$35 = (3)^3 + 3(3) = 1$

$g'(35) = \dfrac{1}{3(3)^2 + 3} = \dfrac{1}{30}$

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