Chapter 4.2, Problem 58E

(a)

To determine

To estimate: The absolute maximum and minimum values (round off up to two decimals) of the function from its graph.

Answer to Problem 58E

The absolute maximum value of $f\left(x\right)$ is $f\left(1\right)\approx 2.85$

The absolute minimum value of $f\left(x\right)$ is $f\left(0.231\right)\approx 1.89$

Explanation of Solution

Given:

The function $f\left(x\right)=e^x+e^{-2x},0\le x\le 1$.

Graph:

Use the graphing calculator to draw the graph of the function $f\left(x\right)=e^x+e^{-2x},0\le x\le 1$ as shown below in Figure 1.

Calculation:

From Figure 1, it is observed that at the point $\left(1,2.85\right)$ the curve attains the maximum value. Therefore, the absolute maximum value occurs at $x=1$ and the value is $2.85$.

Also from Figure 1, it is observed that at the point $\left(0.231,1.89\right)$ the curve attains the minimum value. Therefore, the absolute minimum value occurs at $x=0.231$ and the value is $1.89$.

Thus, the absolute maximum value of $f\left(x\right)$ is $f\left(1\right)\approx 2.85$ and the absolute minimum value of $f\left(x\right)$ is $f\left(0.231\right)\approx 1.89$.

(b)

To determine

To find: The exact maximum and minimum values of the given function $f\left(x\right)=e^x+e^{-2x},0\le x\le 1$.

Answer to Problem 58E

The absolute maximum value of $f\left(x\right)$ is $f\left(1\right)\approx 2.8536$.

The absolute minimum value of $f\left(x\right)$ is $f\left(\ln 2^{\frac{1}{3}}\right)\approx 1.8899$.

Explanation of Solution

Obtain the first derivative of the given function.

$\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{d}{dx}\left(e^x+e^{-2x}\right)\\ =e^x+\left(-2e^{-2x}\right)\\ =e^x-2e^{-2x}\end{array}$

Set $f^{\prime }\left(x\right)=0$ and obtain the critical points.

$\begin{array}{l}{e}^x-2e^{-2x}=0\\ e^x=2e^{-2x}\end{array}$

Take log on both sides,

$\begin{array}{l}\ln e^x=\ln \left(2e^{-2x}\right)\\ \ln e^x=\ln 2+\ln e^{-2x}\left[\because \ln \left(ab\right)=\ln a+\ln b\right]\\ x=\ln 2-2x\\ 3x=\ln 2\end{array}$

Simplify further.

$\begin{array}{l}x=\frac{1}{3}\ln 2\\ x=\ln \left(2^{\frac{1}{3}}\right)\end{array}$

Here, the critical point is $x=\ln \left(2^{\frac{1}{3}}\right)\approx 0.231$.

Hence, all the critical numbers lie in the given interval $0\le x\le 1$.

Apply the extreme values of the given interval and the critical number values in the function $f\left(x\right)$.

When $x=0$,

$\begin{array}{c}f\left(0\right)=e^{\left(0\right)}+e^{-2\left(0\right)}\\ =1+1\\ =2\end{array}$

When $x=1$,

$\begin{array}{c}f\left(1\right)=e^{\left(1\right)}+e^{-2\left(1\right)}\\ =e^1+e^{-2}\\ \approx 2.8536\end{array}$

When $x=\ln \left(2^{\frac{1}{3}}\right)$,

$\begin{array}{c}f\left(\ln 2^{\frac{1}{3}}\right)=e^{\ln 2^{\frac{1}{3}}}+e^{-2\left(\ln 2^{\frac{1}{3}}\right)}\\ =e^{\ln 2^{\frac{1}{3}}}+e^{\ln 2^{\frac{-2}{3}}}\\ =2^{\frac{1}{3}}+2^{\frac{-2}{3}}\\ \approx 1.8899\end{array}$

Recall the fact that the largest functional value is absolute maximum and smallest functional value is absolute minimum. Compare the above values.

Thus, the absolute maximum value of $f\left(x\right)$ is $f\left(1\right)\approx 2.8536$ and the absolute minimum value of $f\left(x\right)$ is $f\left(\ln 2^{\frac{1}{3}}\right)\approx 1.8899$.