The problem is $\displaystyle f(x)=1/x$ and the difference quotient to evaluate is $\displaystyle (f(x)-f(a))/x-a$ the answer in the book is -1/ax. How did they arrive at that conclusion?
Thanks.
Evan
start with...
$\displaystyle \frac{\frac{1}{x}-\frac{1}{a}}{x-a}$
getting a common denominator in the numerator...
$\displaystyle \frac{\frac{a-x}{ax}}{x-a}$
factor a negative out of the denominator...
$\displaystyle \frac{\frac{a-x}{ax}}{-(a-x)}$
cancel the (a-x) and rewrite...
$\displaystyle \frac{-1}{ax}$
The numerator is $\displaystyle \frac{1}{x}-\frac{1}{a}$.
Think about how you would do this: $\displaystyle \frac{1}{2}-\frac{1}{3}$.
To get a common denominator you would multiply the $\displaystyle \frac{1}{2}$ by $\displaystyle \frac{3}{3}$ and the $\displaystyle \frac{1}{3}$ by $\displaystyle \frac{2}{2}$, right?
$\displaystyle \frac{1}{2}-\frac{1}{3} = \frac{1*3}{2*3}-\frac{1*2}{3*2} = \frac{3-2}{6} = \frac{1}{6}$
Same thing here using x and a instead of 2 and 3.
$\displaystyle \frac{1}{x}-\frac{1}{a} = \frac{1*a}{x*a}-\frac{1*x}{a*x} = \frac{a-x}{ax}$