Chapter 4, Problem 82RE

a.

To determine

To find: $\frac{dy}{dx}.$

Answer to Problem 82RE

  $\frac{dy}{dx}=\frac{e^x-e^{-x}}{2}.$

Explanation of Solution

Given information: Given function is $y=\frac{e^x+e^{-x}}{2}.$

Calculation:

  $\begin{array}{l}y=\frac{e^x+e^{-x}}{2}=\frac{e^x}{2}+\frac{e^{-x}}{2}.\\ \frac{dy}{dx}=(1)\frac{e^x}{2}+(-1)\frac{e^x}{2}=\frac{e^x-e^{-x}}{2}.\end{array}$

b.

To determine

To find: $\frac{d^{2}y}{d^{2}x}.$

Answer to Problem 82RE

  $\frac{d^{2}y}{d^{2}x}=\frac{e^x+e^{-x}}{2}.$

Explanation of Solution

Given information: Given function is $y=\frac{e^x+e^{-x}}{2}.$

Calculation:

  $\begin{array}{l}y=\frac{e^x+e^{-x}}{2}=\frac{e^x}{2}+\frac{e^{-x}}{2}.\\ \frac{dy}{dx}=(1)\frac{e^x}{2}+(-1)\frac{e^x}{2}=\frac{e^x-e^{-x}}{2}.\\ \frac{d^{2}y}{d^{2}x}=\frac{e^x-(-1)e^{-x}}{2}=\frac{e^x+e^{-x}}{2}.\end{array}$

c.

To determine

To find: an equation of the line tangent to the curve at x =1.

Answer to Problem 82RE

The equation of tangent line is $y=1.175x+\mathrm{0.368.}$

Explanation of Solution

Given information: Given function is $y=\frac{e^x+e^{-x}}{2}.$

Formula used:

The equation of a line in point slope form is

  $y-y_1=m(x-x_1).$ Where m is the slope and $(x_1,y_1)$ is the point of the line.

Calculation:

  $\begin{array}{l}\frac{dy}{dx}=\frac{e^x-e^{-x}}{2}.\\ $ {(\frac{dy}{dx})}_{x=1}=\frac{e^1-e^{-1}}{2}\approx 1.175.$y(1)=\frac{e^1+e^{-1}}{2}\approx \mathrm{1.543.}\end{array}$

Substitute the point (1,1.543) and slope of m =1.175 into the point-slope form of a line to write the equation of the tangent line.

  $\begin{array}{l}y-1.543=1.175(x-1).\\ y=1.175x+\mathrm{0.368.}\end{array}$

d.

To determine

To find: an equation of the line normal to the curve at x =1.

Answer to Problem 82RE

Equation of the normal line is $y=-0.851x+\mathrm{0.692.}$

Explanation of Solution

Given information: Given function is $y=\frac{e^x+e^{-x}}{2}.$

Calculation:

From part (c ) ,the point is (1,1.543) and the slope of the tangent is 1.175 so the slope of the normal line is then -1/1.175 $\approx -0.851$ since perpendicular slopes are negative reciprocal.

So the equation of the normal line will be,

  $\begin{array}{l}y-1.543=-0.851(x-1).\\ y=-0.851x+\mathrm{0.692.}\end{array}$

e.

To determine

To find: any points where the tangent line is horizontal.

Answer to Problem 82RE

The tangent line is horizontal at x = 0.

Explanation of Solution

Given information: Given function is $y=\frac{e^x+e^{-x}}{2}.$

Calculation:

The tangent line is horizontal when y’=0;

  $\begin{array}{c}y=\frac{e^x+e^{-x}}{2}.\\ y\text{'}=\frac{e^x-e^{-x}}{2}=0.\\ e^x-e^{-x}=0.\\ e^x=e^{-x}.\\ x=0.\end{array}$

So, the tangent line is horizontal at x = 0.