Chapter 4.2, Problem 10E

a.

To determine

To find: the equation of the secant line AB.

Answer to Problem 10E

The equation of the secant line is $y=\frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}$ .

Explanation of Solution

Given:

  $f\left(x\right)=\sqrt{x-1}$ on $1\le x\le 3$ .

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 $x=1$ in $f\left(x\right)=\sqrt{x-1}$ , it follows that

  $\begin{array}{c}f\left(1\right)=\sqrt{1-1}\\ =0\end{array}$

By substituting $x=3$ in $f\left(x\right)=\sqrt{x-1}$ , it follows that

  $\begin{array}{c}f\left(1\right)=\sqrt{3-1}\\ =\sqrt{2}\end{array}$

Now equation of the secant line passing through the points $\left(1,0\right)$ and $\left(3,\sqrt{2}\right)$ is,

  $\begin{array}{c}y-0=\frac{\sqrt{2}-0}{3-1}\left(x-1\right)\\ y=\frac{\sqrt{2}}{2}\left(x-1\right)\\ y=\frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{2}\end{array}$

b.

To determine

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

Answer to Problem 10E

The equation of the tangent line is $y=\frac{\sqrt{2}}{2}x-\frac{\sqrt{2}}{4}$ .

Explanation of Solution

Given:

  $f\left(x\right)=\sqrt{x-1}$ on $1\le x\le 3$ .

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,

  $f\left(x\right)=\sqrt{x-1}$

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

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

Therefore the function is continuous on $\left[1,3\right]$ and differentiable on $\left(1,3\right)$ .

Since the slope of the secant line is $\frac{\sqrt{2}}{2}$ , therefore the slope of the tangent line is also $\frac{\sqrt{2}}{2}$ . That is,

  $\begin{array}{c}{f}^{\prime }\left(x\right)=\frac{\sqrt{2}}{2}\\ \frac{1}{2\sqrt{x-1}}=\frac{\sqrt{2}}{2}\\ \frac{1}{\sqrt{x-1}}=\sqrt{2}\\ \frac{1}{x-1}=2\\ x-1=\frac{1}{2}\\ x=\frac{3}{2}\end{array}$

Now when $x=\frac{3}{2}$ ,

  $\begin{array}{c}f\left(\frac{3}{2}\right)=\sqrt{\frac{3}{2}-1}\\ =\sqrt{\frac{1}{2}}\\ =\frac{\sqrt{2}}{2}\end{array}$

Now the equation of the tangent line is,

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