Question

Given that x) = x^2+x+1, f(-1) = 1" alt="f(x) = x^2+x+1, f(-1) = 1" align="absmiddle" class="latex-formula"> and $g(x)$ is the inverse of $f(x)$, find $g'(1)$
Include an explanation of how you found your answer.

Hint: You will probably want to use implicit differentiation.

Answer 1

Answer:

-1

Step-by-step explanation:

Note that by definition,

$f(g(x))=x \implies f'(g(x)) g'(x) = 1 \implies g'(x)=\frac{1}{f'(g(x))}$

Substituting x=1,

$g'(1)=\frac{1}{f'(g(1))}$

As g is the inverse of f,

$f(-1)=1 \implies g(1)=-1 \\ \\ \implies g'(1)=\frac{1}{f'(-1)}$

By the power rule, f'(x)=2x+1, so f'(-1)=-1.

$\implies g'(1)=\frac{1}{-1}=-1$