1. ## [SOLVED] Inverse Function

I would like to find the inverse function of $f(x)=x^3+3^x$

Since $a^b = \exp{b\ln{a}}$

$y=x^3+\exp{x\ln{3}}$

I do not know how to solve for x in terms of y.

2. You can not solve it for x in terms of y by using the standard methods.
Do you post all informations of this problem?

3. Originally Posted by Paperwings
I would like to find the inverse function of $f(x)=x^3+3^x$

Since $a^b = \exp{b\ln{a}}$

$y=x^3+\exp{x\ln{3}}$

I do not know how to solve for x in terms of y.
Quite simply, you don't- not in terms of elementary functions.

4. Originally Posted by Paperwings
I would like to find the inverse function of $f(x)=x^3+3^x$

Since $a^b = \exp{b\ln{a}}$

$y=x^3+\exp{x\ln{3}}$

I do not know how to solve for x in terms of y.
why do you need the inverse of this function?

is there more top this problem than what you have stated?

5. Let $f(x)=x^3+3^x$. show that f has an inverse and find $(f^{-1})'{(4)}$.

$(f^{-1})'{(c)} = \frac{1}{f'(a)}$.

If f(a) = 4, then a = 1. Since $f'(x)=3x^2+3^x\ln{3}$, f'(1)=3+3ln3

Thus,

$(f^{-1})'{(4)} = \frac{1}{f'(1)} = \frac{1}{3(1+\ln{3})}$.

I was just curious to see what the inverse of f was. I assume that to show that f has an inverse means to solve for the inverse. Since this function is an increasing function, it has an inverse.

6. Originally Posted by Paperwings
$f^{-1}{(4)} = \frac{1}{f'(1)}$.
This is wrong.
I think you mean the DERIVATIVE of the inverse function at 4 not just the value of the inverse function at 4.

7. Originally Posted by General
This is wrong.
I think you mean the DERIVATIVE of the inverse function at 4 not just the value of the inverse function at 4.
Yes, you are correct. I was fixing some Latex syntax errors, and forgot to put the derivative sign. Edit: Fixed it.

8. Since f(1) = 4

f^(-1)(4) = 1

f^(-1) ' (4) = 1/ f '(1) = 1/[3+3ln(2)]