Question

Find the minimum value of the function $9e^x+25e^{-x}$.

Answer 1

Let $f\left(x\right)=$$9e^x+25e^{-x}$.

Now, $f^{\prime }\left(x\right)=9e^x-25e^{-x}$

And $f^{\prime \prime }\left(x\right)=9e^x+25e^{-x}$.

For maximum or minimum value of $f\left(x\right)$ we must have $f^{\prime }\left(x\right)=0$

The we get,

$9e^x=25e^{-x}$

or, $e^{2x}=\frac{25}{9}$

or, $e^x=\frac{5}{3}$ [ Since $e^x>0\forall x$

Now for $e^x=\frac{5}{3}$ we have $f^{\prime \prime }\left(x\right)>0$.

So $e^x=\frac{5}{3}$ gives the minimum value of the function $f\left(x\right)$.

The minimum value of the function $f\left(x\right)$ is $=9.\frac{5}{3}+25.\frac{3}{5}=15+15=30$.