# Interesting Question

• Mar 4th 2006, 12:51 PM
Jameson
Interesting Question
This is from a Mu Alpha Theta Competition I took today.

19. Let f and g be functions such that $\displaystyle f(x)^2=g^{-1}(4x-1)$. What is $\displaystyle f(x)f'(x)g'(f(x)^2)$?

A) Not enough information
B)2
C) $\displaystyle \frac{2}{4x-1}$
D)8x-2

I didn't really know how to approach this, so a little push would be great.
• Mar 4th 2006, 01:23 PM
CaptainBlack
Quote:

Originally Posted by Jameson
This is from a Mu Alpha Theta Competition I took today.

19. Let f and g be functions such that $\displaystyle f(x)^2=g^{-1}(4x-1)$. What is $\displaystyle f(x)f'(x)g'(f(x)^2)$?

A) Not enough information
B)2
C) $\displaystyle \frac{2}{4x-1}$
D)8x-2

I didn't really know how to approach this, so a little push would be great.

$\displaystyle f(x)^2=g^{-1}(4x-1)$

Rewrite as:

$\displaystyle g([f(x)]^2)=4x-1$

Now differentiate wrt x.

RonL
• Mar 4th 2006, 03:05 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
$\displaystyle f(x)^2=g^{-1}(4x-1)$

Rewrite as:

$\displaystyle g([f(x)]^2)=4x-1$

Now differentiate wrt x.

RonL

How do you know it is differenciable :D
• Mar 4th 2006, 10:04 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
How do you know it is differenciable :D

1. It is a linear form in $\displaystyle x$ and so differentiable wrt $\displaystyle x$.

2. If $\displaystyle f$ and $\displaystyle g$ are differentiable then this is differentiable. Asking the
question implies we may assume that $\displaystyle f$ and $\displaystyle g$ are differentiable, just
as it allows us to assume that $\displaystyle '$ denotes the derivative.

RonL
• Mar 5th 2006, 09:58 AM
ThePerfectHacker
I was making a remark that in analysis you always need to show diffrenciability before taking the derivative, I just found that funny.
• Mar 5th 2006, 01:11 PM
Jameson
Quote:

Originally Posted by CaptainBlack
$\displaystyle f(x)^2=g^{-1}(4x-1)$

Rewrite as:

$\displaystyle g([f(x)]^2)=4x-1$

Now differentiate wrt x.

RonL

So I get that $\displaystyle g'[f(x)^2]*2f(x)*f'(x)=4$, so my answer is 2.
• Mar 5th 2006, 01:24 PM
CaptainBlack
Quote:

Originally Posted by Jameson
So I get that $\displaystyle g'[f(x)^2]*2f(x)*f'(x)=4$, so my answer is 2.

That's what I get :)

RonL