Chapter 5.2, Problem 10E

a.

To determine

To find : The equation for the secant line $AB$ .

Answer to Problem 10E

The equation of the secant line $AB$ is $y=\frac{1}{\sqrt{2}}x-\frac{1}{\sqrt{2}}$ .

Explanation of Solution

Given information :

The interval $a\le x\le b$ , $A=\left(a,f\left(a\right)\right)$ , $B=\left(b,f\left(b\right)\right)$ and $f\left(x\right)=\sqrt{x-1}\text{ 1}\le x\le 3$

Calculation :

Since the equation of the line passing through the point $\left(a,f\left(a\right)\right)$ and $\left(b,f\left(b\right)\right)$ is given by

  $y-f\left(a\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\left(x-a\right)$ ..... (1)

Where $\frac{f\left(b\right)-f\left(a\right)}{b-a}$ is slope of the line $AB$ .

Here, $a=1$ and $b=3$ , therefore

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

And

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

Substitute these values in equation (1) and solve:

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

Hence,

The equation of the secant line $AB$ is $y=\frac{1}{\sqrt{2}}x-\frac{1}{\sqrt{2}}$ .

b.

To determine

To find : The equation of tangent line of $f$ in the interval $\left(a,b\right)$ that is parallel to $AB$ .

Answer to Problem 10E

The equation of a tangent that is parallel to $AB$ is $y=\frac{1}{\sqrt{2}}x-\frac{1}{2\sqrt{2}}$ .

Explanation of Solution

Given information :

The interval $a\le x\le b$ , $A=\left(a,f\left(a\right)\right)$ , $B=\left(b,f\left(b\right)\right)$ and $f\left(x\right)=\sqrt{x-1}\text{ 1}\le x\le 3$

Calculation :

Since, the slope of tangent of the $f\left(x\right)=\sqrt{x-1}$ is derivative of the function $f\left(x\right)$ .

Differentiate $f\left(x\right)$ with respect to $x$ :

  $\begin{array}{l}f\left(x\right)=\sqrt{x-1}\\ f\text{'}\left(x\right)=\frac{1}{2\sqrt{x-1}}\end{array}$

Since, the tangent line is parallel to $AB$ , therefore the slope of the tangent will be same as $AB$ .

The slope of $AB$ from (a) is $\frac{1}{\sqrt{2}}$ , therefore substitute $f\text{'}\left(x\right)=\frac{1}{\sqrt{2}}$ in above derivative and solve for $x$ :

  $\begin{array}{l}\frac{1}{\sqrt{2}}=\frac{1}{2\sqrt{x-1}}\\ \sqrt{2}=2\sqrt{x-1}\\ 2=4\left(x-1\right)\\ 2=4x-4\\ 4x=6\\ x=\frac{6}{4}\\ x=\frac{3}{2}\end{array}$

Also,

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

Therefore, the equation of the line will be:

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

Hence,

The equation of a tangent that is parallel to $AB$ is $y=\frac{1}{\sqrt{2}}x-\frac{1}{2\sqrt{2}}$ .