Chapter 4, Problem 61RE

a.

To determine

Find the equation of the line normal to the graph of f at the point where x =2.

Answer to Problem 61RE

  $y=\frac{-1}{3}x+\frac{14}{3}.$

Explanation of Solution

Given information: A differentiable function f has the property that $f\text{'}(x)>0$ for all x .

The line $y=3x-2$ is tangent to the graph of f at the point where x= 2.

Calculation:

  $y=3x-2,(2,4)$

From the given information, we can find out that the intersecting point is at (2,4).

We can find the slope of the normal line by taking the negative reciprocal.

Then find the b and plug it back in the equation.

  $\begin{array}{l}y=\frac{-1}{3}x+b.\\ 4=\frac{-2}{3}+b.\\ b=\frac{14}{3}.\\ So,\\ y=\frac{-1}{3}x+\frac{14}{3}.\end{array}$

b.

To determine

Find the equation of the line tangent to the graph of $f^{-1}$ at the point where y =2.

Answer to Problem 61RE

  $3y=x+2$ .

Explanation of Solution

Given information: A differentiable function f has the property that $f\text{'}(x)>0$ for all x .

The line $y=3x-2$ is tangent to the graph of f at the point where x= 2.

Calculation:

The tangent will have slope 1/3 and will pass through(4,2).So,Equation of the tangent is,

  $\begin{array}{l}y=\frac{x}{3}+c.\\ 2=\frac{4}{3}+c.\\ c=\frac{2}{3}.\\ y=\frac{x}{3}+\frac{2}{3}.\\ 3y=x+2.\end{array}$

c.

To determine

Find the equation of the line tangent to the graph of $y=\frac{f(x)}{x}$ at the point where x =2.

Answer to Problem 61RE

  $y=\frac{1}{2}x+1.$

Explanation of Solution

Given information: A differentiable function f has the property that $f\text{'}(x)>0$ for all x .

The line $y=3x-2$ is tangent to the graph of f at the point where x= 2.

Calculation:

  $\begin{array}{l}y=\frac{f(x)}{x}.\\ y(2)=\frac{f(2)}{2}=\frac{4}{2}=2.(2,2)\\ y\text{'}(2)=\frac{2f\text{'}(2)-f(2)}{4}=\frac{2(3)-4}{4}=\frac{1}{2}.\\ y-2=\frac{1}{2}(x-2).\\ y=\frac{1}{2}x+1.\end{array}$