Question #dc111

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Answer 1 · 2016-09-30T17:49:02

1

Since g(x) is inverse of f(x), f(g(x)=x

Differentiating both sides w.r.t x, it would be

f'(g(x)) g'(x) =1

Thus for x=0, f'(g(0))g'(0)=1

Answer 2 · 2016-09-30T18:09:19

The function is not invertible (it does not have an inverse).
$f(1) = f(-1)$ So $f$ is not one-to-one. Ignoring that for the moment, we can get $1$ .

We can use the result:

If $g$ is the inverse of $f$ , then $g'(c) = 1/(f'(g(c))$

Therefore, if there were any such thing as an inverse of the function given, we would have

$f'(g(0))g'(0) = f'(g(0)) * 1/(f'(g(0))) = 1$