# Math Help - difference quotient and absolute value

1. ## difference quotient and absolute value

"Evaluate the difference quotient for the given functions. Show all of your work."

The ones I am stuck on are:

a. f(x) = lxl if x<-1 and 0<h<1

b. f(x) = lxl if x>1 and 0<h<1

I know the answers -- a. -1, b. 1 -- but I got them by plugging numbers that fit the parameters into the difference quotient and then evaluating it, which would be fine if I didn't have to show my work. As far as evaluating it directly, thought, I have no idea what to do. (My latex skills are pathetic, so please excuse my not typing all of my work out.)

2. Originally Posted by sinewave85
"Evaluate the difference quotient for the given functions. Show all of your work."

The ones I am stuck on are:

a. f(x) = lxl if x<-1 and 0<h<1

b. f(x) = lxl if x>1 and 0<h<1

I know the answers -- a. -1, b. 1 -- but I got them by plugging numbers that fit the parameters into the difference quotient and then evaluating it, which would be fine if I didn't have to show my work. As far as evaluating it directly, thought, I have no idea what to do. (My latex skills are pathetic, so please excuse my not typing all of my work out.)
Let's do a):

$\frac{|x + h| - |x|}{h}$

Now, x is less than -1 and 0 < h < 1 we know that x + h is negative. Thus $|x + h| = -(x + h)$. Again, since x is negative $|x| = -x$. Thus
$\frac{|x + h| - |x|}{h} = \frac{-(x + h) - (-x)}{h} = \frac{-x - h + x}{h} = \frac{-h}{h} = -1$

-Dan

3. Originally Posted by topsquark
Now, x is less than -1 and 0 < h < 1 we know that x + h is negative. Thus $|x + h| = -(x + h)$. Again, since x is negative $|x| = -x$. Thus

$\frac{|x + h| - |x|}{h} = \frac{-(x + h) - (-x)}{h} = \frac{-x - h + x}{h} = \frac{-h}{h} = -1$

-Dan
Oh, wow. I never really understood absolute value correctly untill just now. That is both amazing and sad. Oh, well. Glad I decided to do calculus, or I would have gone my whole life not understanding absolute value. Just sad. Thanks for the enlightenment.