# Thread: Evaluate and simplfy this expression

1. ## Evaluate and simplfy this expression

The question is:

Evaluate and simplify the expression

[ f(x+h)-f(x) ] / h ; h =/= 0

for f(x) = 1/(x+3).

I know how to plug in the function into the expression...
but for some reason I don't remember where to go from there.
How would you simplify it?

Thanks

2. You have $\displaystyle f'(x) = \displaystyle \lim_{h\to{0}}\frac{\frac{1}{x+h+3}-\frac{1}{x+3}}{h}$

Let $\displaystyle \frac{1}{x+h+3} = a$ and $\displaystyle \frac{1}{x+3} = b$, then $\displaystyle h = \frac{1}{a}-\frac{1}{b}$, so:

$\displaystyle \displaystyle f'(x) = \lim_{a\to{b}}\frac{a-b}{\frac{1}{a}-\frac{1}{b}} = \lim_{a\to{b}}\frac{ab(a-b)}{b-a}= -\lim_{a\to{b}}\frac{ab(a-b)}{a-b} = \displaystyle = -\lim_{a\to{b}}ab = -b^2 = -\frac{1}{(x+3)^2}$

3. Whoa.
Uh.
Sorry, maybe I didn't specify, but I just need to know how to algebraically simplify the expression.
I think you used calculus in there somewhere...
hahaha I have no idea what that says.

This might sound stupid, but I just need to know how to simplify the fractions once you plug in f(x) into the expression.
I'm having this massive brain fart the night before the test hahaha..

thanks

4. Originally Posted by jellyksong
Sorry, maybe I didn't specify, but I just need to know how to algebraically simplify the expression.
Well, that's what I'd done, and it's what the question asks.

Anyway, is this what you wanted?

$\displaystyle \displaystyle \frac{\frac{1}{x+h+3}-\frac{1}{x+3}}{h} = \frac{\frac{(x+3)-(x+h+3)}{(x+3)(x+h+3)}}{h} = \frac{(x+3)-(x+h+3)}{h(x+3)(x+h+3)}$

$\displaystyle \displaystyle = \frac{-h}{h(x+3)(x+h+3)} = \frac{-1}{(x+3)(x+h+3)}$

5. OH, I get it now hahaha
you have to make a common denominator...

Thank you so much!