Chapter 3, Problem 61RE

(a).

To determine

To find: The equation of line normal to the graph of $f$ at the point where $x=2$ .

Answer to Problem 61RE

The equation of the line normal to the graph is:

  $y=4-\frac{1}{3}\left(x-2\right)$

Explanation of Solution

Given:

The equation of line $y=3x-2$

Concept Used:

• Equation of line normal to the graph is:

$y-y_0=-\frac{1}{m}\left(x-x_0\right)$

• Slope of normal= $\frac{-1}{\frac{dy}{dx}}$

Calculation:

Differentiation is as follow:

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left[3x-2\right]\\ =3\end{array}$

  $\begin{array}{c}\text{Slope}\text{of}\text{normal}=\frac{-1}{{\left[\frac{dy}{dx}\right]}_{x=2}}\\ =\frac{-1}{3}\end{array}$

  $\begin{array}{c}x=2\\ y=3\left(2\right)-2\\ =4\end{array}$

The points are: $\left(2,4\right)$ .

Conclusion:

The equation of the line normal to the graph is:

  $y=4-\frac{1}{3}\left(x-2\right)$

(b).

To determine

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

Answer to Problem 61RE

The tangent line has the equation:

  $\left(y-2\right)=\frac{1}{3}\left(x-4\right)$

Explanation of Solution

Given:

The equation of line $y=3x-2$

Concept Used:

According to the inverse function theorem, if $f$ is a differentiable function with nonzero derivative at the point $a$ then:

  $\left(f^{-1}\right)\text{'}\left(f\left(a\right)\right)=\frac{1}{f\text{'}\left(a\right)}$

Calculation:

The points are: $\left(4,2\right)$ .

The slope of the tangent line of

  $f^{-1}\left(x\right)\text{at}\left(4,2\right)$ .

This can also be written as:

  $f\left(4\right)=2$ .

By inverse function theorem:

  $\begin{array}{c}\left(f^{-1}\right)\text{'}\left(4\right)=\left(f^{-1}\right)\left(f\left(2\right)\right)\\ =\frac{1}{f\text{'}\left(2\right)}\\ =\frac{1}{3}\end{array}$

The tangent line has the equation:

  $\left(y-2\right)=\frac{1}{3}\left(x-4\right)$

Conclusion:

The tangent line has the equation:

  $\left(y-2\right)=\frac{1}{3}\left(x-4\right)$

(c).

To determine

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

Answer to Problem 61RE

The equation of line tangent to the graph is:

  $\begin{array}{c}y=\frac{1}{x}\left[2+3\left(x-2\right)\right]\\ =3\end{array}$

Explanation of Solution

Given:

• The equation of line $y=3x-2$
• $y=\frac{f\left(x\right)}{x}$

Concept Used:

• Equation of line normal to the graph is:

$y-y_0=m\left(x-x_0\right)$

• Slope of normal= $\frac{dy}{dx}$

Calculation:

  $y=3-\frac{2}{x}$

The points are:

  $x=2,y=2$

The points are: $\left(2,2\right)$ .

Then, slope is:

  $\frac{dy}{dx}=\frac{2}{x^2}$

  $\begin{array}{c}\text{slope}=\frac{2}{4}\\ =\frac{1}{2}\end{array}$

Conclusion:

The equation of line tangent to the graph is:

  $y=2+\frac{1}{2}\left(x-2\right)$