Question #dc111

2 Answers
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

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#.

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