# Thread: Finding the derivative of an inverse function

1. ## Finding the derivative of an inverse function

Find $(f^{-1})(a)$ for the function $f$ and real number $a$.

$f(x)=x^3+2x-1$, $a=2$

$f(x)=\frac{1}{27}(x^5+2x^3)$, $a=-11$

I'm having trouble with these. I've got a test on monday and I'm freaking out. The biggest problem that I'm having is finding the number that corresponds to a. I can't sub for $f(x)$ because the function can't be solved easily in most cases, and I can't determine the inverse function by switching variables and solving for y. What do I do?

2. for first, f(1)=2....hit and trial

3. for second, f(-3)=-11.......again hit and trial

4. Inverse Function Theorem

$(f^{-1})'(b) = \frac {1}{f'(a)}$ where b =f(a)

5. Originally Posted by malaygoel
for first, f(1)=2....hit and trial

Hit and trial is the only way to go huh? There's gotta be something I'm missing here.

6. $f(x) = x^{3}+2x-1$

$f'(x)= 3x^{2}+2$

$(f^{-1})'\big(f(x)\big) = \frac {1}{f'(x)}$

$(f^{-1})'(x^3+2x-1) = \frac {1}{3x^{2}+2}$

$(f^{-1})'(2) = (f^{-1})'(1^{3}+2(1)-1) = \frac {1}{3(1^{2})+2}$

so $(f^{-1})'(2) = \frac {1}{5}$

7. Originally Posted by Random Variable
$f(x) = x^{3}+2x-1$

$f'(x)= 3x^{2}+2$

$(f^{-1})'\big(f(x)\big) = \frac {1}{f'(x)}$

$(f^{-1})'(x^3+2x-1) = \frac {1}{3x^{2}+2}$

$(f^{-1})'(2) = (f^{-1})'(1^{3}+2(1)-1) = \frac {1}{3(1^{2})+2}$

so $(f^{-1})'(2) = \frac {1}{5}$
I know how to evaluate once I have the value. The question is: How do I find the value? Some things simply can't be solved. Do I just try inspection? Malaygoel seems to think that if a is an integer, then inspection is the easiest way. Is he right?

8. Originally Posted by VonNemo19
I know how to evaluate once I have the value. The question is: How do I find the value? Some things simply can't be solved. Do I just try inspection? Malaygoel seems to think that if a is an integer, then inspection is the easiest way. Is he right?
in general, by inspection (trial and error), is the most efficient way to find the inverse value, and it is usually the way you are expected to do it. most of the time, professors will use functions where it is hard to compute the value, but it is easily seen.

here is how it would be computed. we wish to find $f^{-1}(a)$, so call this value $x$ (this is a suggestive name). so we have that

$f^{-1}(a) = x$

by taking $f$ of both sides, we obtain the equation

$a = f(x)$

and so we see, we are looking for a value that when we plug it into our function, the result is $a$. this is what the inverse function dos after all, gives you the original input that gave you the current output, namely $a$.

so for example, to deal with your first problem, we would find $f^{-1}(a)$ by solving the equation

$x^3 + 2x - 1 = 2$, that is

$x^3 + 2x - 3 = 0$

now this is not that hard to solve, however, it takes a lot less time and effort to just eye-ball the solution at the beginning as malaygoel did