Chapter 4.2, Problem 9E

a.

To determine

To find: the equation of the secant line AB.

Answer to Problem 9E

The equation of the secant line is $y=2.5$ .

Explanation of Solution

Given:

  $f\left(x\right)=x+\frac{1}{x}$ on $0.5\le x\le 2$ .

Concept used:

Equation of a line passing through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$ .

Calculation:

By substituting $x=0.5$ in $f\left(x\right)=x+\frac{1}{x}$ , it follows that

  $\begin{array}{c}f\left(0.5\right)=0.5+\frac{1}{0.5}\\ =2.5\end{array}$

By substituting $x=2$ in $f\left(x\right)=x+\frac{1}{x}$ , it follows that

  $\begin{array}{c}f\left(2\right)=2+\frac{1}{2}\\ =2.5\end{array}$

Now equation of the secant line passing through the points $\left(0.5,2.5\right)$ and $\left(2,2.5\right)$ is,

  $\begin{array}{c}y-2.5=\frac{2.5-2.5}{2-0.5}\left(x-0.5\right)\\ y-2.5=0\\ y=2.5\end{array}$

b.

To determine

To find: the equation of the tangent line parallel to the secant line.

Answer to Problem 9E

The equation of the tangent line is $y=2$ .

Explanation of Solution

Given:

  $f\left(x\right)=x+\frac{1}{x}$ on $0.5\le x\le 2$ .

Concept used:

Mean Value Theorem for Derivatives:

If $y=f\left(x\right)$ is continuous at every point of the closed interval $\left[a,b\right]$ and differentiable at every point of its interior $\left(a,b\right)$ , then there is at least one point $c$ in $\left(a,b\right)$ at which the instantaneous rate of change equals the mean rate of change,

  $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Calculation:

Consider the given function,

  $f\left(x\right)=x+\frac{1}{x}$

By differentiating with respect to $x$ , it follows that

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

Therefore the function is continuous on $\left[0.5,2\right]$ and differentiable on $\left(0.5,2\right)$ .

Since $f\left(0.5\right)=f\left(2\right)=2.5$ , therefore the tangent is horizontal.

So the slope of the tangent is 0. That is,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=0\\ 1-\frac{1}{x^2}=0\\ \frac{1}{x^2}=1\\ x^2=1\\ x=1\end{array}$

Now the equation of the tangent line is,

  $\begin{array}{l}y=1+\frac{1}{1}\\ y=2\end{array}$