# Consider the function defined as f(x) = e^x + e^-x/2, x > 0. Find the inverse of f?

Apr 18, 2018

Assuming $f \left(x\right) = \frac{{e}^{x} + {e}^{-} x}{2}$, then

${f}^{-} 1 \left(x\right) = \ln \left(x + \sqrt{{x}^{2} - 1}\right)$.

But if $f \left(x\right) = {e}^{x} + {e}^{-} \frac{x}{2}$, then

${f}^{-} 1 \left(x\right) = \ln \left(\frac{x + \sqrt{{x}^{2} - 2}}{2}\right)$

#### Explanation:

I'm not sure if you wrote the function correctly. You wrote

$f \left(x\right) = {e}^{x} + {e}^{-} \frac{x}{2}$

Did you mean to find the inverse of

$f \left(x\right) = \frac{{e}^{x} + {e}^{-} x}{2}$?

$f \left(x\right) = y = \frac{{e}^{x} + {e}^{-} x}{2}$

Multiply both sides by 2.

$2 y = {e}^{x} + {e}^{-} x$

Multiply both sides by ${e}^{x}$.

$2 y {e}^{x} = {\left({e}^{x}\right)}^{2} + 1$

Subtract $2 y {e}^{x}$ from both sides.

$0 = {\left({e}^{x}\right)}^{2} - 2 y {e}^{x} + 1$

Use the quadratic formula to solve for ${e}^{x}$.

${e}^{x} = y \pm \sqrt{{y}^{2} - 1}$

But ${e}^{x}$ and $y$ are all positive so we can just write

${e}^{x} = y + \sqrt{{y}^{2} - 1}$

Now take the natural log of both sides.

$x = \ln \left(y + \sqrt{{y}^{2} - 1}\right)$

Another way to say this is

${f}^{-} 1 \left(x\right) = \ln \left(x + \sqrt{{x}^{2} - 1}\right)$.

If you really DID mean $f \left(x\right) = {e}^{x} + {e}^{-} \frac{x}{2}$, then the answer is

${f}^{-} 1 \left(x\right) = \ln \left(\frac{x + \sqrt{{x}^{2} - 2}}{2}\right)$

using exactly the same method as above.