1. evaluating difference quotients

The problem is $f(x)=1/x$ and the difference quotient to evaluate is $(f(x)-f(a))/x-a$ the answer in the book is -1/ax. How did they arrive at that conclusion?

Thanks.
Evan

2. show what have you done.

3. Originally Posted by Evan.Kimia
The problem is $f(x)=1/x$ and the difference quotient to evaluate is $(f(x)-f(a))/x-a$ the answer in the book is -1/ax. How did they arrive at that conclusion?
$\frac{\frac{1}{x}-\frac{1}{a}}{x-a}$
getting a common denominator in the numerator...
$\frac{\frac{a-x}{ax}}{x-a}$
factor a negative out of the denominator...
$\frac{\frac{a-x}{ax}}{-(a-x)}$
cancel the (a-x) and rewrite...
$\frac{-1}{ax}$

4. i plugged in f(x) and f(a) into the equation and got...

$(1/x)-(1/a)/(x-a)$

im not sure how to simplify this numerator correctly.

5. Originally Posted by pflo
$\frac{\frac{1}{x}-\frac{1}{a}}{x-a}$
getting a common denominator in the numerator...
$\frac{\frac{a-x}{ax}}{x-a}$
factor a negative out of the denominator...
$\frac{\frac{a-x}{ax}}{-(a-x)}$
cancel the (a-x) and rewrite...
$\frac{-1}{ax}$
I see that you factored out a-x by -1 in order to cancel out the a-x above, but what method would i use to get a common denominator in the numerator?

Thanks.

6. Originally Posted by Evan.Kimia
I see that you factored out a-x by -1 in order to cancel out the a-x above, but what method would i use to get a common denominator in the numerator?
The numerator is $\frac{1}{x}-\frac{1}{a}$.

Think about how you would do this: $\frac{1}{2}-\frac{1}{3}$.

To get a common denominator you would multiply the $\frac{1}{2}$ by $\frac{3}{3}$ and the $\frac{1}{3}$ by $\frac{2}{2}$, right?

$\frac{1}{2}-\frac{1}{3} = \frac{1*3}{2*3}-\frac{1*2}{3*2} = \frac{3-2}{6} = \frac{1}{6}$

Same thing here using x and a instead of 2 and 3.

$\frac{1}{x}-\frac{1}{a} = \frac{1*a}{x*a}-\frac{1*x}{a*x} = \frac{a-x}{ax}$

7. Thank you so much everyone.