Chapter 1.3, Problem 39E

Solution

a.

To Determine: The interval in which the function is increasing or decreasing.

The function is increasing on the interval of $\left(0,\infty \right)$ .

The function is decreasing on the interval of $\left(-\infty ,0\right)$ .

Given information:

  $h\left(x\right)=\left|x\right|-10$

Graph:

  

From the graph, it can be seen that the function $h\left(x\right)=\left|x\right|-10$ is increasing on the interval of $\left(0,\infty \right)$ and decreasing on the interval of $\left(-\infty ,0\right)$ .

b.

To Determine: The function is odd, even or neither.

The function is even.

Given information:

  $h\left(x\right)=\left|x\right|-10$

Calculation:

A function is even if $f\left(-x\right)=f\left(x\right)$

Check if $f\left(-x\right)=f\left(x\right)$ :

  $\left|x\right|-10=\left|x\right|-10$

Since $\left|x\right|-10=\left|x\right|-10$ , the function is even.

So, the function $h\left(x\right)=\left|x\right|-10$ is even.

c.

To Determine: The extrema of the function, if any.

The function has absolute minimum at $\left(0,-10\right)$ .

Given information:

  $h\left(x\right)=\left|x\right|-10$

Calculation:

Find the first derivative of the function.

  $h^{\text{'}}\left(x\right)=\frac{x}{\left|x\right|}$

Find the second derivative of the function.

  $h^{\prime \prime }\left(x\right)=0$

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

Set the first derivative equal to $0$ :

  $\frac{x}{\left|x\right|}=0$

Set the numerator equal to zero.

  $x=0$

Exclude the solutions that do not make $\frac{x}{\left|x\right|}=0$ true, so there is no solution.

Set the denominator in $\left|x\right|$ equal to $0$ to find where the expression is undefined:

  $\begin{array}{c}\left|x\right|=0\\ x=\pm 0\\ x=0\end{array}$

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

Split $\left(-\infty ,\infty \right)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

  $\left(-\infty ,0\right)\cup \left(0,\infty \right)$

Substitute any number, such as $-2$ , from the interval $\left(-\infty ,0\right)$ in the first derivative to check if the result is negative or positive.

  $\begin{array}{l}{h}^{\prime }\left(-2\right)=\frac{-2}{|-2|}\\ h^{\prime }\left(-2\right)=-1\end{array}$

Substitute any number, such as $2$ , from the interval $\left(0,\infty \right)$ in the first derivative $\frac{x}{\left|x\right|}$ to check if the result is negative or positive:

  $\begin{array}{l}{h}^{\prime }\left(2\right)=\frac{2}{|-2|}\\ h^{\prime }\left(2\right)=1\end{array}$

Since the first derivative changed signs from negative to positive around $x=0$ . This function has local minimum at $x=0$ .

  

And from the graph, it can be seen that the function $h\left(x\right)=\left|x\right|-10$ has absolute minimum at $\left(0,-10\right)$ .

d.

To Determine: The graph of the function related to a graph of one of the twelve basic functions.

The graph is related to absolute value function.

Given information:

  $h\left(x\right)=\left|x\right|-10$

Calculation:

  

From the above graph, it can be seen that the function $h\left(x\right)=\left|x\right|-10$ is absolute value function, since it shifted 10 units down.