# How do you find the exact minimum value of f(x) = e^x + e^(-2x) on [0,1]?

Aug 23, 2015

See the explanation

#### Explanation:

The minimum is either $f \left(0\right) = 2$

or $f \left(1\right) = e + {e}^{-} 2 = e + \frac{1}{e} ^ 2 = \frac{{e}^{3} + 1}{e} ^ 2$

or $f \left(c\right)$ for some critical number of $f$ in $\left[0 , 1\right]$.

So we need to find critical numbers for $f$

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

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

$= \frac{{e}^{3 x} - 2}{e} ^ \left(2 x\right)$

Since the denominator ${e}^{2 x}$ is never $0$ (it is always positive), we see that $f '$ is never undefined.

Solving $f ' \left(x\right) = 0$ gets us ${e}^{3 x} = 2$ so

$3 x = \ln 2$ and $x = \ln \frac{2}{3} = \ln \sqrt[3]{2}$

Because $\ln$ is an increasing function, and $1 < \sqrt[3]{2} < e$,

we see that $\ln \sqrt[3]{2}$ is in $\left[0 , 1\right]$

So $c = \ln \sqrt[3]{2} = \ln \frac{2}{3}$ is the only critical number.

We need to determine whether $f \left(c\right)$ is a minimum, a maximum or neither.

The denominator of $f ' \left(x\right)$ is always positive, so the sign of $f ' \left(x\right)$ will be the same as the sign of the numerator:

${e}^{3 x} - 2$.

For $x < \ln \frac{2}{3}$, we have $3 x < \ln 2$, and ${e}^{3 x} < {e}^{\ln} 2 = 2$

so ${e}^{3 x} - 2$. and $f ' \left(x\right)$ is negative.

By similar reasoning, for $x > \ln \frac{2}{3}$, we have $f ' \left(x\right) > 0$

Therefore $f \left(\ln \frac{2}{3}\right)$ is a local minimum.

Furthermore, $f \left(\ln \frac{2}{3}\right) < f \left(0\right)$ since $f$ is decreasing on $\left[0 , \ln \frac{2}{3}\right)$
and $f \left(\ln \frac{2}{3}\right) < f \left(1\right)$ since $f$ is increasing on $\left(\ln \frac{2}{3} , 1\right]$

So on the interval $\left[0 , 1\right]$ the value $f \left(\ln \frac{2}{3}\right)$ is the absolute minimum.

f(lnroot(3)2) = e^(lnroot(3)2) + e^(-2(ln(2^(1/3)))

$= {e}^{\ln \sqrt[3]{2}} + {e}^{\ln} \left({2}^{- \frac{2}{3}}\right)$

$= \sqrt[3]{2} + \frac{1}{\sqrt[3]{4}}$

The minimum value of $f \left(x\right) = {e}^{x} + {e}^{- 2 x}$ on [0,1] is $\sqrt[3]{2} + \frac{1}{\sqrt[3]{4}}$

(Rewrite using algebra until you're happy with the way it looks. Personally, I like: $\frac{3}{\sqrt[3]{4}}$)

Now that we're finished, it might be nice to see the graph of $f$

graph{(y - e^x-e^(-2x))=0 [-0.965, 2.453, 1.486, 3.195]}