Chapter 4, Problem 62RE

a.

To determine

Find the equation of the line normal to the graph of g at the point where x =1.

Answer to Problem 62RE

  $y=\frac{1}{2}x+\frac{5}{2}.$

Explanation of Solution

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

The line $y=5-2x$ is tangent to the graph of g at the point where x= 1.

Calculation:

  $\begin{array}{l}y=5-2x.\\ x=1.\\ y=5-2(1)=5-2=3.\end{array}$ .

The slope of the tangent line $y=5-2x$ is -2 so the slop of the normal line is m=1/2 since perpendicular slopes are negative reciprocals. So the equation of the line passes through (1,3) and slop=1/3 is,

  $\begin{array}{l}y-y_1=m(x-x_1)\\ y-3=\frac{1}{2}(x-1).\\ y=\frac{1}{2}x-\frac{1}{2}+3.\\ y=\frac{1}{2}x+\frac{5}{2}.\end{array}$

b.

To determine

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

Answer to Problem 62RE

  $y=-\frac{1}{2}x+\frac{5}{2}.$

Explanation of Solution

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

The line $y=5-2x$ is tangent to the graph of g at the point where x= 1.

Calculation:

  $\begin{array}{l}y=5-2x.\\ x=1.\\ y=5-2(1)=5-2=3.\end{array}$

The tangent line then passes through the point (1,3) on g ,then $g^{-1}$ has the point(3,1).The slop of the tangent line $y=5-2x$ for g is -2 .Then the slop of $g^{-1}$ is -1/2.

So, tangent on $g^{-1}$ passes through (3,1) and has slop=-1/2.So the equation of the tangent on $g^{-1}$ is,

  $\begin{array}{l}y-1=-\frac{1}{2}(x-3).\\ y=-\frac{1}{2}x+\frac{3}{2}+1.\\ y=-\frac{1}{2}x+\frac{5}{2}.\end{array}$

c.

To determine

Find the equation of the line tangent to the graph of $y=g(x^2)$ at the point where x =1.

Answer to Problem 62RE

  $y=-4x+7.$

Explanation of Solution

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

The line $y=5-2x$ is tangent to the graph of g at the point where x= 1.

Calculation:

  $\begin{array}{l}g(x)=5-2x.\\ g(1)=5-2(1)=5-2=3.\\ y=g(x^2).\\ x=1.\\ y=g(1^2)=g(1)=3(1,3)\\ y=g(x^2).\\ y\text{'}=g\text{'}(x^2).2x.\\ y\text{'}(1)=2.(1).g\text{'}(1^2)\\ y\text{'}(1)=2g\text{'}(1)\\ y\text{'}(1)=2(-2)\\ y\text{'}(1)=-4.\end{array}$

Find $y\text{'}$ and then evaluate the derivative at x =1 to find the slop of the tangent line for $y=g(x^2)$ when x =1.Since the slop of the tangent line of g is -2 then $g\text{'}(1)=-2$

Equation of the line tangent to the graph of $y=g(x^2)$ is

  $\begin{array}{l}y-3=-4(x-1)\\ y=-4x+7.\end{array}$