Learn / Library / Calculus

Bending of an inverse

CalculusHard

Let $f (x) = x^{3} + 2 x + 1$ , which is increasing on the whole real line, and let $g$ be its inverse function. Find $g^{''} (4)$ . Give the answer as a fraction in lowest terms.

Checking an answer needs JavaScript. The hint and the solution open without it.

Hint

Find which $x$ maps to $4$ , then differentiate the identity $f (g (y)) = y$ twice.

Solution

Since $f (1) = 1 + 2 + 1 = 4$ , we have $g (4) = 1$ .
Differentiating $f (g (y)) = y$ gives $g^{'} (y) = \frac{1}{f ^{'} ( g ( y ))}$ , and $f^{'} (x) = 3 x^{2} + 2$ , so $g^{'} (4) = \frac{1}{5}$ .
Differentiating again gives $g^{''} (y) = - \frac{f ^{''} ( x )}{( f ^{'} ( x ) ) ^{3}}$ with $x = g (y)$ .
Here $f^{''} (x) = 6 x$ , so $f^{''} (1) = 6$ and $f^{'} (1) = 5$ .
Therefore $g^{''} (4) = - \frac{6}{5 \times 5 \times 5} = - \frac{6}{125}$ .

Answer: $- \frac{6}{125}$

Next question Back to the library