Proving existence of inverse function using the rule of differentiating inverse functions.

Question

I am working from a textbook that gives us that

The derivative of the inverse function is $$ \frac{dx}{dy} = \large{\frac{1}{\frac{dy}{dx}}}.\qquad (*)$$

$$$$ Now I am stuck on part $(a)$ of the following question:

A function is defined by $\ f(x) =x^3 + 3x + 2.$

$(a)$ By considering $f'(x),\ $ prove that $\ f(x)\ $ has an inverse function.

$(b)\ $ Find the gradient of the graph of $y=f^{-1}(x)$ at the point where $x=2.$

$$$$ For $(b),\ f(x) = 2\ \iff x=0\ $. Therefore, the gradient of the graph of $y=f^{-1}(x)$ at the point where $\ x=2\ $ is equal to $\ \large{\frac{1}{f'(0)}} = \frac{1}{3}.$

But I'm simply a bit confused unsure on what it wants for part $(a)$... The derivative of the inverse function is $\ \large{\frac{1}{3}\cdot \frac{1}{x^2+1} },\ $ but this does not say that, for example, the derivative of the inverse function $\ y=f^{-1}(x)\ $ at the point $\ x=3\ $ is $\ \large{\frac{1}{3}\cdot \frac{1}{3^2+1} },\ $ which it isn't. This means I find $(*)$ confusing as to what it is saying exactly, and like I say, I'm confused about this and not sure how to tackle part $(a)$. Please can someone clarify. Thanks.

Answer 1

$(a)\quad f'(x) =3(x^2 +1)\geq3>0\ $ for all $\ x.\ $ Therefore $\ f(x)\ $ is strictly increasing, and therefore one-to-one. Since a one-to-one function has an inverse, $\ f(x)\ $ has an inverse, i.e. $\ f^{-1}(x)\ $ exists.

Answer 2

The derivative of an inverse can be written more clearly as $$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))},$$ and part $(a)$ can be shown straightforwardly using the inverse function theorem (https://en.wikipedia.org/wiki/Inverse_function_theorem).

Answer 3

(a) Let g(.) be the inverse function of f(.)

f'(x) =  3(x^2 + 1) > 0 for all x.

f is strictly increasing, therefore, it is one to one.

Hence its inverse g: R-->R, defined by g(x) = $f^{-1}(x)$

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

  By definition g(f(x)) = x.

  Differentiating wrt x, by chain rule of differentiation,  we have

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

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

(b) g'(f(2)) = $\frac{1} {3(2^2 + 1)}$ = $\frac{1} {15}$