Question

Tangent Lines

1. If $g\left(x\right)=\frac{1}{2x-1},$find $g\text{'}\left(a\right).$
2. Find equations of tangent lines to the graph of $g$at the points whose $x-coordinates$are $-1,0,\&1.$

c. Graph $g$and the three tangent lines..

Short answer

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

(b) The equation of tangent lines:

$\begin{array}{l}y=-\frac{2}{9}x-\frac{5}{9}\\ y=-2x-1\\ y=-2x+3\end{array}$.

(c) Graph of g and three tangent lines is,

Step-by-step solution

Step 1. Given information.

The function here given is,

$g\left(x\right)=\frac{1}{2x-1}$

Step 2. Formula used.

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

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

if this limit exists.

Step 3. Finding the slope of the tangent line at a given point.

According to the definition of a derivative at a, we have

$\begin{array}{c}g\text{'}\left(a\right)=\underset{h\to 0}{\lim }\frac{f(a+h)-f\left(a\right)}{h}Definitionoff\text{'}\left(a\right)\\ =\underset{h\to 0}{\lim }\frac{\left(\frac{1}{2(a+h)-1}\right)-\left(\frac{1}{2a-1}\right)}{h}\text{g}\left(x\right)=\frac{1}{2x-1}\\ =\underset{h\to 0}{\lim }\\ =\underset{h\to 0}{\lim }\frac{2a-1-2a-2h+1}{h\left(\left(2\right(a+h)-1)(2a-1)\right)}Simplify\\ =\underset{h\to 0}{\lim }-\frac{2\cancel{h}}{\cancel{h}\left(\left(2\right(a+h)-1)(2a-1)\right)}Cancelh\\ =-\frac{2}{\left(\left(2\right(a+0)-1)(2a-1)\right)}\\ =-\frac{1}{{(2a-1)}^2}\end{array}$

Therefore, the derivative of the function at is .

To find the slope of tangent line at given point,

We have.

Let’s find the value of derivative at .

Substitute the value of into the

The slopes of the tangents at are respectively.