Chapter 13.3, Problem 26E

a.

To determine

To find : $g\text{'}\left(a\right)$if $g\left(x\right)=\frac{1}{2x-1}$.

Answer to Problem 26E

  $g\text{'}\left(a\right)=\frac{-2}{{\left(2a-1\right)}^2}$

Explanation of Solution

Given information:

The given function is $g\left(x\right)=\frac{1}{2x-1}$.

Calculation :

The derivative of a function $f$ at a number $a$ , denoted by $f\text{'}\left(a\right)$ ,is

  $f\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$

provided this limit exist.

According to definition, $g\text{'}\left(a\right)$ is

  $\begin{array}{l}g\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}\\ g\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{\frac{1}{2\left(a+h\right)-1}-\frac{1}{\left(2a-1\right)}}{h}\\ g\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{\frac{\left(2a-1\right)-2\left(a+h\right)+1}{2\left(a+h\right)\left(2a-1\right)-1\left(2a-1\right)}}{h}\\ g\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{2a-1-2a-2h+1}{h\left[2\left(a+h\right)\left(2a-1\right)-1\left(2a-1\right)\right]}\\ g\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{-2\cancel{h}}{\cancel{h}\left[2\left(a+h\right)\left(2a-1\right)-1\left(2a-1\right)\right]}\\ g\text{'}\left(a\right)=\frac{-2}{2\left(a+0\right)\left(2a-1\right)-1\left(2a-1\right)}\\ g\text{'}\left(a\right)=\frac{-2}{{\left(2a-1\right)}^2}\end{array}$

Hence,

  $g\text{'}\left(a\right)=\frac{-2}{{\left(2a-1\right)}^2}$

b.

To determine

To find : Equation of the tangent lines to the graph g at the points whose x-coordinate are $-1$ , 0, and 1.

Answer to Problem 26E

The equation of tangents whose x coordinate are $-1$ , 0, and 1 are $y=-\frac{2}{9}x-\frac{5}{9}$ , $y=-2x-1$ and $y=-2x+3$.

Explanation of Solution

Given information:

The function is $g\left(x\right)=\frac{1}{2x-1}$ and $g\text{'}\left(a\right)=\frac{-2}{{\left(2a-1\right)}^2}$

Calculation :

The equation of a tangent line to the graph of $g$ at $x=a$ is the line passing through the point $\left(a,g\left(a\right)\right)$ , whose slope is $g\text{'}\left(a\right)$ . Therefore,

At $x=-1$ , tangent line is

  $y-g\left(-1\right)=g\text{'}\left(-1\right)\left(x+1\right)$

Here,

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

  $\begin{array}{l}g\text{'}\left(-1\right)=-\frac{2}{{\left[2\left(-1\right)-1\right]}^2}\\ g\text{'}\left(-1\right)=-\frac{2}{9}\end{array}$

Thus the equation of the tangent at $x=-1$ is

  $\begin{array}{l}y-\left(-\frac{1}{3}\right)=-\frac{2}{9}\left(x+1\right)\\ y=-\frac{2}{9}x-\frac{5}{9}\end{array}$

Similarly, at $x=0$ , tangent line is

  $y-g\left(0\right)=g\text{'}\left(0\right)\left(x-0\right)$

Here,

  $\begin{array}{l}g\left(0\right)=\frac{1}{2\cdot 0-1}\\ g\left(0\right)=-1\end{array}$ and

  $\begin{array}{l}g\text{'}\left(0\right)=-\frac{2}{{\left[2\left(0\right)-1\right]}^2}\\ g\text{'}\left(0\right)=-2\end{array}$

Thus the equation of the tangent at $x=0$ is

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

And, at $x=1$ , tangent line is

  $y-g\left(1\right)=g\text{'}\left(1\right)\left(x-1\right)$

Here,

  $\begin{array}{l}g\left(1\right)=\frac{1}{2\cdot 1-1}\\ g\left(1\right)=1\end{array}$ and

  $\begin{array}{l}g\text{'}\left(1\right)=-\frac{2}{{\left(2\cdot 1-1\right)}^2}\\ g\text{'}\left(1\right)=-2\end{array}$

Thus the equation of the tangent at $x=1$ is

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

Hence,

The equation of tangents whose x coordinate are $-1$ , 0, and 1 are $y=-\frac{2}{9}x-\frac{5}{9}$ , $y=-2x-1$ and $y=-2x+3$.

c.

To determine

To graph : graph f and the three tangent lines.

Answer to Problem 26E

The graph is drawn below.

Explanation of Solution

Given information:

The function is $g\left(x\right)=\frac{1}{2x-1}$ and three tangents are $y=-\frac{2}{9}x-\frac{5}{9}$ , $y=-2x-1$ and $y=-2x+3$ .

Calculation :

The graph of the function f and the three tangent lines is drawn below: