Chapter 3.2, Problem 33E

(A)

To determine

To find:

The equation of the tangent line to the curve at the given point

Answer to Problem 33E

The equation of the tangent line is $y=\frac{-2x\left(x+1\right)}{{\left(x^2+1\right)}^2}+\frac{1}{2}$

Explanation of Solution

Given:

The function is

  $y=\frac{1}{x^2+1},\left(-1,\frac{1}{2}\right)$

Concept used:

An equation of a line $l$ through a point $P_1\left(x_1,y_1\right)$ with slope $m$ . If $P\left(x,y\right)$ is any point with $x\ne x_1$

Then P is on $l$ if and only if the slope of the line through $P_1$ and P

The equation is $y-y_1=m\left(x-x_1\right)$

Calculation:

The function

  $y=\frac{1}{x^2+1}\mathrm{..................}\left(1\right)$

The derivative of a function

  $\begin{array}{l}y=f\left(x\right)\\ y^{\prime }=f^{\prime }\left(x\right)=\frac{dy}{dx}\end{array}$

Differentiating the equation (1) with respect to $x$

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{x^2+1}\right)\\ \frac{dy}{dx}=\frac{d}{dx}{\left(x^2+1\right)}^{-1}\\ m=\frac{dy}{dx}=\frac{-2x}{{\left(x^2+1\right)}^2}\end{array}$

The point is $\left(-1,\frac{1}{2}\right)$

The equation is

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-\frac{1}{2}=\left(\frac{-2x}{{\left(x^2+1\right)}^2}\right)\left(x+1\right)\\ y=\frac{-2x\left(x+1\right)}{{\left(x^2+1\right)}^2}+\frac{1}{2}\end{array}$

(B)

To determine

To draw:

The graph of the tangent line to the curve at the given point

Explanation of Solution

Given:

The function is

  $y=\frac{1}{x^2+1},\left(-1,\frac{1}{2}\right)$

Concept used:

An equation of a line $l$ through a point $P_1\left(x_1,y_1\right)$ with slope $m$ . If $P\left(x,y\right)$ is any point with $x\ne x_1$

Then P is on $l$ if and only if the slope of the line through $P_1$ and P

The equation is $y-y_1=m\left(x-x_1\right)$

Calculation:

The function

  $y=\frac{1}{x^2+1}\mathrm{..................}\left(1\right)$

The derivative of a function

  $\begin{array}{l}y=f\left(x\right)\\ y^{\prime }=f^{\prime }\left(x\right)=\frac{dy}{dx}\end{array}$

Differentiating the equation (1) with respect to $x$

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{x^2+1}\right)\\ \frac{dy}{dx}=\frac{d}{dx}{\left(x^2+1\right)}^{-1}\\ m=\frac{dy}{dx}=\frac{-2x}{{\left(x^2+1\right)}^2}\end{array}$

The point is $\left(-1,\frac{1}{2}\right)$

The equation is

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-\frac{1}{2}=\left(\frac{-2x}{{\left(x^2+1\right)}^2}\right)\left(x+1\right)\\ y=\frac{-2x\left(x+1\right)}{{\left(x^2+1\right)}^2}+\frac{1}{2}\end{array}$

Draw the table

  $f\left(x\right)=\frac{1}{x^2+1},\left(-1,\frac{1}{2}\right)$

Test one point in each of the region formed by the graph

If the point satisfies the function then shade the entire region to denote that every point in the region satisfies the function

$x-axis$	$0$	$-1$	$-2$	$2$
$y-axis$	$1$	$0.5$	$0.2$	$0.2$