1. ## inverse function question

the question is....Let f(x)=x^3+x. If h is the inverse function of f, then h'(2)=

the answer is 1/4, but I dont understand how to do it..this problem is from the collegeboard clep guide and it doesn't explain any problems.

Thanks for any help I can get

2. Originally Posted by jst706
the question is....Let f(x)=x^3+x. If h is the inverse function of f, then h'(2)=

the answer is 1/4, but I dont understand how to do it..this problem is from the collegeboard clep guide and it doesn't explain any problems.

Thanks for any help I can get
use the formula: $\displaystyle \left( f^{-1} \right)'(x) = \frac 1{f' \left( f^{-1}(x) \right)}$

here, $\displaystyle h(x) = f^{-1}(x)$

3. every time I try that I get a negative answer, I must be doing something wrong

4. Originally Posted by jst706
every time I try that I get a negative answer, I must be doing something wrong
do you mind showing your work, so that we can pin-point the areas where you are making mistakes?

5. Maybe I don't understand the formula...this is what I have been doing.

x^-3+x^-1
-3x^-4-x^-2

= 1/ (-3x^4-x^2)

h'(2)=1/(-3(16)-4)

6. Originally Posted by jst706
Maybe I don't understand the formula...
it seems you are correct

this is what I have been doing.

no, that is NOT what you should do. that is finding $\displaystyle \left( \frac 1{f(x)} \right)'$ not $\displaystyle \frac 1{f'(x)}$

x^-3+x^-1
-3x^-4-x^-2

= 1/ (-3x^4-x^2)
even if you were correct to do this, this is still wrong. this is not how you find the derivative of a fraction

ok, let's take this slow.

the formula is: $\displaystyle h'(x) = \frac 1{f'(h(x))}$

we want $\displaystyle h'(2)$, it means we must find $\displaystyle h(2)$ and plug it into $\displaystyle f'(x)$ (this, of course, means we also have to find $\displaystyle f'(x)$)to get $\displaystyle f'(h(2))$ and then plug it into the formula.

first we find $\displaystyle f'(x)$:

$\displaystyle f'(x) = 3x^2 + 1$

ok, that wasn't so hard.

now let's find $\displaystyle h(2)$. since $\displaystyle h(x) = f^{-1}(x)$, this amounts to finding an $\displaystyle x$ such that $\displaystyle f(x) = 2$

thus we must solve: $\displaystyle x^3 + x = 2$

it is almost obvious that $\displaystyle x = 1$

thus, $\displaystyle h(2) = 1$

this means that $\displaystyle f'(h(2)) = f'(1) = 4$.

now we just plug that into our formula:

$\displaystyle h'(2) = \frac 1{f'(h(2))} = \frac 14$