# Inverses and Second Derivatives

• Sep 9th 2009, 05:59 AM
uberbandgeek6
Inverses and Second Derivatives
Problem:
If f is a one-to-one, twice differentiable function with inverse function g, show that:

g''(x) = -( f''(g(x)) ) / ( [f'(g(x))]^3 )

and deduce that if f is increasing and concave upward, then its inverse function is concave downward.
___________________________

My work so far:
For the first part, I replaced all of the g's with f^-1 (I won't show that here since it would be really cluttered).
Then I took the equation (f^-1)'(x) = 1 / (f'(f^-1(x))) and changed it to
f'(f^-1(x)) = 1 / ((f^-1)'(x)) and plugged that into the bottom of the problem's equation. After a little simplifying, I got
(f^-1)''(x) = (-f''(f^-1(x))) * ((f^-1)'(x))^3

I'm not sure where to go from here. What can I do about the second derivatives? Thank you!
• Sep 9th 2009, 07:54 AM
hatsoff
Quote:

Originally Posted by uberbandgeek6
Problem:
If f is a one-to-one, twice differentiable function with inverse function g, show that:

g''(x) = -( f''(g(x)) ) / ( [f'(g(x))]^3 )

and deduce that if f is increasing and concave upward, then its inverse function is concave downward.
___________________________

My work so far:
For the first part, I replaced all of the g's with f^-1 (I won't show that here since it would be really cluttered).
Then I took the equation (f^-1)'(x) = 1 / (f'(f^-1(x))) and changed it to
f'(f^-1(x)) = 1 / ((f^-1)'(x)) and plugged that into the bottom of the problem's equation. After a little simplifying, I got
(f^-1)''(x) = (-f''(f^-1(x))) * ((f^-1)'(x))^3

I'm not sure where to go from here. What can I do about the second derivatives? Thank you!

Part I

Remember the identity that for inverses, we have

$f[g(x)]=x$.

Taking the derivative, we get

$f'[g(x)]g'(x)=1$.

Divide by $f'g$ to get:

$g'(x)=(f'[g(x)])^{-1}$.

Take the derivative again, yielding

$g''(x)=(-1)(f'[g(x)])^{-2}f''[g(x)]g'(x)$.

Remember that $g'(x)=(f'[g(x)])^{-1}$ and substitute:

$g''(x)=(-1)(f'[g(x)])^{-2}f''[g(x)](f'[g(x)])^{-1}$.

And then simplify to yield

$g''(x)=-\frac{f''[g(x)]}{(f'[g(x)])^{3}}$.

Part II

Now, when we say that $f$ is increasing and concave upward, we are essentially saying that

$f'(t),f''(t)>0$ for all $t\in D$. Since $g(x)\in D$, then

$f'[g(x)],f''[g(x)]>0$.

We can use that information with our previous equations to infer that $g''(x)<0$.