Chapter 12.3, Problem 44E

a.

To determine

Find the slope of the graph.

Answer to Problem 44E

  $\boxed{-2}$ .

Explanation of Solution

Given information:

Find the slope of the graph of $f$ at a given point.

  $f(x)=4-x^2$ , $(1,3)$

Calculation:

Consider the function, $f(x)=4-x^2$ .

The slope of the graph $f(x)$ of at a point $(x,y)$

  $f\text{'}(x)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f(x)}{h}$

So, the slope of the graph $f(x)=4-x^2$ is,

  $f\text{'}(x)=\underset{h\to 0}{\lim }\frac{4-{\left(x+h\right)}^2-4+x^2}{h}$

Use mathematical identity $\left(a+b\right)=a^2+2ab+b^2$

  $\begin{array}{l}f\text{'}(x)=\underset{h\to 0}{\lim }\frac{4-x^2-h^2-2xh-4+x^2}{h}\\ =\underset{h\to 0}{\lim }(h-2x)\\ =-2x\\ f\text{'}\left(x\right)=-2\end{array}$

At point $(1,3)$ , the slope will be ,

  $f\text{'}\left(x\right)=-2$

Hence, the slope of the graph $f(x)=4-x^2$ at $(1,3)$ is $\boxed{-2}$ .

b.

To determine

Find an equaion of the tangent line to the graph at the point.

Answer to Problem 44E

  $\boxed{y=-2x+5}$

Explanation of Solution

Given information:

Find an equaion of the tangent line to the graph at the point.

  $f(x)=4-x^2$ , $(1,3)$

Calculation:

Since the equation of a standard line is given by,

  $y=mx+c$

Where $m$ is the slope of the line and $c$ is the $y$ intercept made by the line.

Slope of $f(x)=4-x^2$ is $-2$ .

So,

Since the equation of the line is to be found on point $(1,3)$ ,so it should satisfy the line $y=-2x+c$

Now, put the values, $x=1$ and $y=3$ in the equation $y=-2x+c$

  $3=-2\times 1+c$

  $c=5$

Hence, the equation of the tangent line at a point $(1,3)$ of the graph $f(x)=4-x^2$ is

  $\boxed{y=-2x+5}$ .

c.

To determine

Graph the function and the tangent line.

Answer to Problem 44E

  

Explanation of Solution

Given information:

Graph the function and the tangent line.

  $f(x)=4-x^2$ , $(1,3)$

Calculation:

Use maple to draw the graph of the function $f(x)=4-x^2$ and the tangent line $y=-2x+5$ can be drawn as below,

  

Hence, the result shown in graph.