Chapter 11.3, Problem 26E

To determine

To find: the slopes at a given points

Answer to Problem 26E

The slopes at a given points are $A=-\frac{1}{4}andB=-1$

Explanation of Solution

Given:

  $\begin{array}{l}f(x)=\frac{1}{x-2}\\ (a)\left(0,\frac{1}{2}\right)\\ \left(b\right)\left(-3,-1\right)\end{array}$

Calculation:

This is the given equation

  $f(x)=\frac{1}{x-2}$

Use this equation top find the derivative, which gives us the slope of the tangent line

  $m=\underset{h\to 0}{\lim }\frac{f(x+h)-F(x)}{h}$

Now plug in $f(x)$ into the derivative finding equation

  $\underset{h\to 0}{\lim }\frac{\frac{1}{x+h+2}-\frac{1}{x+2}}{h}$

To get rid of the fractions in the numerator, multiply by the LCD

  $\underset{h\to 0}{\lim }\frac{\left(\frac{1}{x+h+2}-\frac{1}{x+2}\right)(x+h+2)(x+2)}{h(x+h+2)(x+2)}$

This is the equation after doing some distribution

  $\underset{h\to 0}{\lim }\frac{(x+2)-(x+h+2)}{h(x+h+2)(x+2)}$

This is after combining like terms

  $\underset{h\to 0}{\lim }\frac{-h}{h(x+2)(x+h+2)}$

The h's cancel out and leave us with this

  $\underset{h\to 0}{\lim }\frac{-1}{(x+2)(x+h+2)}$

Next plug in 0 for h since that is what the limit for the variable h is

  $\frac{-1}{(x+2)(x+0+2)}$

This is the derivative

  $\frac{-1}{{(x+2)}^2}$

Next plug in the $\text{x}$ part of the coordinate from a and b individually

This is a

  $\frac{-1}{{(0+2)}^2}=-\frac{1}{4}$

This is b

  $\frac{-1}{{(-3+2)}^2}=-1$

Thus, $A=-\frac{1}{4}andB=-1$

Conclusion:

Therefore, the slopes at a given points are $A=-\frac{1}{4}andB=-1$