Hello everyone, I'm stuck on this question:

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

Any help appreciated, thx

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- Jul 13th 2013, 08:59 PMJellyOnionFinding 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 - Jul 13th 2013, 09:18 PMMarkFLRe: Finding the min value
Are you sure you have the expression typed correctly? As given it is $\displaystyle 3e^x$, and as such has no extrema.

- Jul 13th 2013, 09:21 PMJellyOnionRe: Finding the min value
Exactly, that's what I thought. My textbook says the answer is -0.5log(e)^2

- Jul 13th 2013, 09:46 PMMarkFLRe: Finding the min value
In that case, if I am interpreting the given critical value correctly, the function should be:

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

Hence, differentiating and equating to zero, we find:

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

This implies:

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

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

Converting from exponential to logarithmic form, we obtain:

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

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