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Question

If g is the inverse function of f and f(x)=x3+3x1f(x)=x^3+3 x-1, what is the value of g'(35)?

Solution

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Given that gg is the inverse of ff and

f(x)=x3+3x1f(x) = x^3 + 3x - 1

The function ff 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 gg is the inverse of ff, we have

g(x)=f1(x)f(g(x))=xddxf(g(x))=ddxxf(g(x))g(x)=1g(x)=1f(g(x))\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 ff

f(x)=x3+3x1f(x)=3x2+3\begin{align*} f(x) & = x^3 + 3x - 1\\ f'(x) & = 3x^2 + 3 \end{align*}

Substituting in the equation for the derivative of gg gives us

g(x)=13(g(x))2+3              substitutedg(35)=13(g(35))2+3             evaluated\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 gg is the inverse function of f,f, we can say that g(35)g(35) is the value of xx for which f(x)=35f(x) = 35. Thus, we need to solve

35=x3+3x135 = x^3 + 3x - 1

Usually we would have to solve this cubic for xx. but the solution is apparent by inspection. We can see that xx has to equal 33.

35=(3)3+3(3)=135 = (3)^3 + 3(3) = 1

So, our answer is

g(35)=13(3)2+3=130g'(35) = \dfrac{1}{3(3)^2 + 3} = \dfrac{1}{30}

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