Hello everyone, I'm stuck on this question:
The minimum value of e^x+2e^x occurs where x = ?
Any help appreciated, thx
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:
$\displaystyle 2x=\ln\left(\frac{1}{2} \right)=-\ln(2)$
$\displaystyle x=-\frac{\ln(2)}{2}$