Chapter 9.3, Problem 9QR

a.

To determine

To find local extreme values of the function and where they occur.

Answer to Problem 9QR

Local minimum at (0, 1) and local maximum at $\approx \left(2,1.54\right)$

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{e^x+x^2}{e^x}$

Calculation:

Take derivative of the function

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

Equate the derivative of the function to 0.

  $\begin{array}{l}\frac{e^x\left(2x-x^2\right)}{{\left(e^x\right)}^2}=0\\ e^x\left(2x-x^2\right)=0\left[\text{multiply each side by}{\left(e^x\right)}^2\right]\\ e^x\cdot x\cdot \left(2-x\right)=0\left[\text{factorise}\right]\\ \text{critial numbers are }0,2\end{array}$

In the interval $(-\infty ,0]$ $f^{\prime }\left(x\right)$ is negative, in the interval $\left[0,2\right]$ $f^{\prime }\left(x\right)$ is negative and in the interval $[2,\infty )$ $f^{\prime }\left(x\right)$ is positive.

Hence, local minimum of the function occurs at 0 and local maximum occurs at 2.

Local minimum value

  $\begin{array}{l}f\left(0\right)=\frac{e^0+0^2}{e^0}\\ f^{\prime }\left(0\right)=1\left[\text{simplify}\right]\end{array}$

Local maximum value

  $\begin{array}{l}f\left(2\right)=\frac{e^2+2^2}{e^2}\\ f^{\prime }\left(2\right)=1.54\left[\text{simplify}\right]\end{array}$

Therefore,

Local minimum at (0,1) and local maximum at $\approx \left(2,1.54\right)$

b.

To determine

To find intervals on which the function is increasing.

Answer to Problem 9QR

The function is increasing in the interval $\left[0,2\right]$

Explanation of Solution

Given information:

The function is $f\left(x\right)=\frac{e^x+x^2}{e^x}$

Calculation:

In the interval $(-\infty ,0]$ $f^{\prime }\left(x\right)$ is negative, in the interval $\left[0,2\right]$ $f^{\prime }\left(x\right)$ is negative and in the interval $[2,\infty )$ $f^{\prime }\left(x\right)$ is positive.

Therefore,

The function is increasing in the interval $\left[0,2\right]$