Chapter 11.1, Problem 20E

Solution

a)

To find:the slope of the graph of the function at the indicated point using the limit of definition of the derivative.

The slope of the graph of the function at the point $x=1$ is $-\frac{1}{9}$ .

Given information:

  $f\left(x\right)=\frac{1}{x+2}$ at $x=1$ .

Formula used:

The derivative of the function $f$ at $x=a$ is donated by $f\text{'}\left(a\right)$ .

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

The derivative of a function at a point can be viewed geometrically as the slope of the line tangent to the curve $y=f\left(x\right)$ .

Calculation:

Calculate the derivative of the function $f$ at $x=t$ is donated by $f\text{'}\left(t\right)$ .

  $\begin{array}{c}f\text{'}\left(x\right)=\underset{h\to 0}{{\displaystyle \lim }}\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ =\underset{h\to 0}{{\displaystyle \lim }}\frac{\frac{1}{\left(x+h\right)+2}-\frac{1}{x+2}}{h}\\ =\underset{h\to 0}{{\displaystyle \lim }}\frac{\left(x+2\right)-\left(x+2+h\right)}{h\left(x+2\right)\left(x+2+h\right)}\\ =-\frac{1}{{\left(x+2\right)}^2}\end{array}$

Calculate $f\text{'}\left(1\right)$ .

  $\begin{array}{c}f\text{'}\left(1\right)=-\frac{1}{{\left(1+2\right)}^2}\\ =-\frac{1}{9}\end{array}$

Therefore,the slope of the graph of the function at the point $x=1$ is $-\frac{1}{9}$ .

b)

To find: an equation of the tangent line at that point.

The equation of the tangent line at point $\left(1,\frac{1}{3}\right)$ and slope $\frac{-1}{9}$ is $y-\frac{1}{3}=-\frac{1}{9}\left(x-1\right)$ .

Given information:

  $f\left(x\right)=\frac{1}{x+2}$ at $x=1$ .

Formula used:

Let $\left(x_1,y_1\right)$ and $m$ are the point on a line and slope of the line respectively, then the equation of line in point slope form is determined by:

  $y-y_1=m\left(x-x_1\right)$

Calculation:

It is given that $f\left(x\right)=\frac{1}{x+2}$ at $x=1$ .

Calculate $f\left(1\right)$ .

  $f\left(1\right)=\frac{1}{1+2}=\frac{1}{3}$

The tangent is passing through point $\left(x,f\left(x\right)\right)$ that is $\left(1,\frac{1}{3}\right)$ . The slope of the line is $\frac{-1}{9}$ as calculated in part (a).

Compute the equation of the tangent line at point $\left(1,\frac{1}{3}\right)$ and slope $-\frac{1}{9}$ .

  $y-\frac{1}{3}=-\frac{1}{9}\left(x-1\right)$

Therefore, the equation of the tangent line at point $\left(1,\frac{1}{3}\right)$ and slope $\frac{-1}{9}$ is $y-\frac{1}{3}=-\frac{1}{9}\left(x-1\right)$ .

c)

To sketch:a graph of the curve near the point without using a grapher.

The graph is a curve like structure.

Given information:

  $f\left(x\right)=\frac{1}{x+2}$ at $x=1$ .

Formula used:

The derivative of the function $f$ at $x=a$ is donated by $f\text{'}\left(a\right)$ .

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

The derivative of a function at a point can be viewed geometrically as the slope of the line tangent to the curve $y=f\left(x\right)$ .

Calculation:

Sketch the graph of $f\left(x\right)=\frac{1}{x+2}$ and point $\left(1,\frac{1}{3}\right)$ .

Prepare a table of points for $f\left(x\right)=\frac{1}{x+2}$ to graph it.

$x$	$f\left(x\right)$
$-1$	$1$
$0$	$\frac{1}{2}$
$1$	$\frac{1}{3}$
$2$	$\frac{1}{4}$

Plot the above points and connect them.

  

Therefore, it can be observed that the graph is a curve like structure.