Question

Let f be the function defined by f(x) = –x^3 – x. If g represents the inverse of f, what is g′(10)?

Answer 1

`\displaystyle g'(10)=(1)/(-13)=-(1)/(13)`

Explanation:

The Derivative of the Inverse Function

Let f(x) be a real invertible function, and g(x) the inverse function of f(x), i.e.,:

`g(x)=f^(-1)(x)`

We can calculate the derivative of the inverse function even if we don't have the inverse function explicitly computed. We use the formula:

`\displaystyle g'(x)=(1)/(f'(g(x)))`

We need to find the value of g'(10) when:

`f(x)=-x^3-x`

Substituting:

`\displaystyle g'(10)=(1)/(f'(g(10)))`

We don't have the value of g(10) but we can guess its value since the inverse functions f and g satisfy:

if y=f(x), then g(y)=x, thus we need to find a value of x that produces a value of f(x)=10.

We can easily find that x=-2:

`f(-2)=-(-2)^3-(-2)=8+2=10.`

Thus, g(10)=-2

Now we find:

`f'(x)=-3x^2-1`

`f'(-2)=-3(-2)^2-1`

`f'(-2)=-3*4-1=-13`

Thus, finally:

`\displaystyle g'(10)=(1)/(-13)=-(1)/(13)`