Question dc111

Sep 30, 2016

1

Explanation:

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

Sep 30, 2016

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

Explanation:

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

$f ' \left(g \left(0\right)\right) g ' \left(0\right) = f ' \left(g \left(0\right)\right) \cdot \frac{1}{f ' \left(g \left(0\right)\right)} = 1$