[SOLVED] Inverse Function

**Paperwings** · January 14th 2010, 06:52 AM

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.

**General** · January 14th 2010, 06:55 AM

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

**HallsofIvy** · January 14th 2010, 06:55 AM

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.

**skeeter** · January 14th 2010, 06:55 AM

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?

**Paperwings** · January 14th 2010, 07:17 AM

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.

**General** · January 14th 2010, 07:47 AM

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.

**Paperwings** · January 14th 2010, 08:01 AM

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.

**Calculus26** · January 14th 2010, 08:03 AM

Since f(1) = 4

f^(-1)(4) = 1

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

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