Chapter 3.1, Problem 60E

(a)

To determine

To find : $\left(f+g\right)\left(-3\right)$ .

Answer to Problem 60E

  $\left(f+g\right)\left(-3\right)=-3.5$

Explanation of Solution

Given information :

The graph of $f$ and, $g$ are

   and respectively.

Formula needed :

  $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$

Substitute $x=-3$ in $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$ .

  $\left(f+g\right)\left(-3\right)=f\left(-3\right)+g\left(-3\right)$

On observing the graph of $f$ at $x=-3$ is $-1.5$ .

On observing the graph of $g$ at $x=-3$ is $-2$ .

  $\begin{array}{c}\left(f+g\right)\left(-3\right)=\left(-1.5\right)+\left(-2\right)\\ =-1.5-2\\ =-3.5\end{array}$

(b)

To determine

To find : ${\left(f+g\right)}^{\prime }\left(-3\right)$ .

Answer to Problem 60E

  ${\left(f+g\right)}^{\prime }\left(-3\right)=1.5$

Explanation of Solution

Given information :

The graph of $f$ and, $g$ are

   and respectively.

Formula needed :

  ${\left(f+g\right)}^{\prime }\left(x\right)=f^{\prime }\left(x\right)+g^{\prime }\left(x\right)$

Substitute $x=-3$ in ${\left(f+g\right)}^{\prime }\left(x\right)=f^{\prime }\left(x\right)+g^{\prime }\left(x\right)$ .

  ${\left(f+g\right)}^{\prime }\left(-3\right)=f^{\prime }\left(-3\right)+g^{\prime }\left(-3\right)$

The $f^{\prime }\left(-3\right)$ is the slope of $f$ at $x=-3$ and $g^{\prime }\left(-3\right)$ is the slope of $g$ at $x=-3$ .

The slope of $f$ at $x=-3$ is $\frac{3-0}{0+2}=1.5$

The slope of $g$ at $x=-3$ is $\frac{0-0}{-2+3}=0$

  $\begin{array}{c}{\left(f+g\right)}^{\prime }\left(-3\right)=1.5+0\\ =1.5\end{array}$

(c)

To determine

To find : $\left(fg\right)\left(1\right)$ .

Answer to Problem 60E

  $\left(fg\right)\left(1\right)=-1.25$

Explanation of Solution

Given information :

The graph of $f$ and, $g$ are

   and respectively.

Formula needed :

  $\left(fg\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$

Substitute $x=1$ in $\left(fg\right)\left(x\right)=f\left(x\right)\cdot g\left(x\right)$ .

  $\left(fg\right)\left(1\right)=f\left(1\right)\cdot g\left(1\right)$

On observing the graph of $f$ at $x=1$ is $2.5$ .

On observing the graph of $g$ at $x=1$ is $-0.5$ .

  $\begin{array}{c}\left(fg\right)\left(1\right)=f\left(1\right)\cdot g\left(1\right)\\ =\left(2.5\right)\cdot \left(-0.5\right)\\ =-1.25\end{array}$

(d)

To determine

To find : ${\left(fg\right)}^{\prime }\left(1\right)$ .

Answer to Problem 60E

  ${\left(fg\right)}^{\prime }\left(1\right)=4$

Explanation of Solution

Given information :

The graph of $f$ and, $g$ are

   and respectively.

Formula needed :

  ${\left(fg\right)}^{\prime }\left(x\right)=f^{\prime }\left(x\right)g\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)$

Substitute $x=1$ in ${\left(fg\right)}^{\prime }\left(x\right)=f^{\prime }\left(x\right)g\left(x\right)+f\left(x\right)g^{\prime }\left(x\right)$ .

  ${\left(fg\right)}^{\prime }\left(1\right)=f^{\prime }\left(1\right)g\left(1\right)+f\left(1\right)g^{\prime }\left(1\right)$

From above, $f\left(1\right)=\left(2.5\right),g\left(1\right)=\left(-0.5\right)$

The $f^{\prime }\left(1\right)$ is the slope of $f$ at $x=1$ and $g^{\prime }\left(1\right)$ is the slope of $g$ at $x=1$ .

The slope of $f$ at $x=1$ is $\frac{2-3}{2-0}=-0.5$ . That is $f^{\prime }\left(1\right)=-0.5$ .

The slope of $g$ at $x=1$ is $\frac{1+2}{2-0}=1.5$ . That is $g^{\prime }\left(1\right)=1.5$ .

Substitute value of $f\left(1\right),f^{\prime }\left(1\right),g\left(1\right),g^{\prime }\left(1\right)$ in ${\left(fg\right)}^{\prime }\left(1\right)=f^{\prime }\left(1\right)g\left(1\right)+f\left(1\right)g^{\prime }\left(1\right)$ .

  $\begin{array}{c}{\left(fg\right)}^{\prime }\left(1\right)=\left(-0.5\right)\left(-0.5\right)+\left(2.5\right)\left(1.5\right)\\ =0.25+3.75\\ =4\end{array}$

(e)

To determine

To find : $\left(f\circ g\right)\left(2\right)$ .

Answer to Problem 60E

  $\left(f\circ g\right)\left(2\right)=2.5$

Explanation of Solution

Given information :

The graph of $f$ and, $g$ are

   and respectively.

Formula needed :

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

Substitute $x=2$ in $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ .

  $\left(f\circ g\right)\left(2\right)=f\left(g\left(2\right)\right)$

On observing the graph of $g$ at $x=2$ is $1$ .

On observing the graph of $f$ at $x=1$ is $2.5$ .

  $\begin{array}{c}\left(f\circ g\right)\left(2\right)=f\left(g\left(2\right)\right)\\ =f\left(1\right)\\ =2.5\end{array}$

(f)

To determine

To find : ${\left(f\circ g\right)}^{\prime }\left(2\right)$ .

Answer to Problem 60E

  ${\left(f\circ g\right)}^{\prime }\left(2\right)=-0.75$

Explanation of Solution

Given information :

The graph of $f$ and, $g$ are

   and respectively.

Formula needed :

  ${\left(f\circ g\right)}^{\prime }\left(x\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)$

Substitute $x=2$ in ${\left(f\circ g\right)}^{\prime }\left(x\right)=f^{\prime }\left(g\left(x\right)\right)\cdot g^{\prime }\left(x\right)$ .

  ${\left(f\circ g\right)}^{\prime }\left(2\right)=f^{\prime }\left(g\left(2\right)\right)\cdot g^{\prime }\left(2\right)$

On observing the graph of $g$ at $x=2$ is $1$ .

  ${\left(f\circ g\right)}^{\prime }\left(2\right)=f^{\prime }\left(1\right)\cdot g^{\prime }\left(2\right)$

From above, $f^{\prime }\left(1\right)=-0.5$ and $g^{\prime }\left(2\right)=1.5$ .

  $\begin{array}{c}{\left(f\circ g\right)}^{\prime }\left(2\right)=\left(-0.5\right)\cdot \left(1.5\right)\\ =-0.75\end{array}$