# Math Help - Finding the min value

1. ## Finding the min value

Hello everyone, I'm stuck on this question:

The minimum value of e^x+2e^x occurs where x = ?

Any help appreciated, thx

2. ## Re: Finding the min value

Are you sure you have the expression typed correctly? As given it is $3e^x$, and as such has no extrema.

3. ## Re: Finding the min value

Exactly, that's what I thought. My textbook says the answer is -0.5log(e)^2

4. ## Re: Finding the min value

In that case, if I am interpreting the given critical value correctly, the function should be:

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

Hence, differentiating and equating to zero, we find:

$\frac{dy}{dx}=-e^{-x}+2e^x=\frac{2e^{2x}-1}{e^x}=0$

This implies:

$2e^{2x}-1=0$

$e^{2x}=\frac{1}{2}$

Converting from exponential to logarithmic form, we obtain:

$2x=\ln\left(\frac{1}{2} \right)=-\ln(2)$

$x=-\frac{\ln(2)}{2}$