# Math Help - Rate of change?

1. ## Rate of change?

Hey guys, here's another one.. I can't believe I don't remember how to do any of this..Thanks.
http://oi44.tinypic.com/33dbec4.jpg

2. ## Re: Rate of change?

$f(x) = \frac{x}{x^2+2}$

determine $\frac{f(x+h)-f(x)}{h}$ in the simplest form

$f(x+h) = \frac{x+h}{(x+h)^2+2}$

and $f(x) = \frac{x}{x^2+2}$
so

$\frac{f(x+h)-f(x)}{h} = \dfrac{\dfrac{x+h}{(x+h)^2+2} - \dfrac{x}{x^2+2}}{h}$

try to cancel the h in the denominator

3. ## Re: Rate of change?

If $f(x)= \frac{x}{x^2+2}$ then $f(x+h)= \frac{x+h}{(x+h)^2+2}$

Now go get em!

4. ## Re: Rate of change?

You guys are the best. Thanks so much! I'll post the answers when I get them

5. ## Re: Rate of change?

Originally Posted by Amer
$f(x) = \frac{x}{x^2+2}$

determine $\frac{f(x+h)-f(x)}{h}$ in the simplest form

$f(x+h) = \frac{x+h}{(x+h)^2+2}$

and $f(h) = \frac{h}{h^2+2}$
so

$\frac{f(x+h)-f(x)}{h} = \dfrac{\dfrac{x+h}{(x+h)^2+2} - \dfrac{h}{h^2+2}}{h}$

try to cancel the h in the denominator
What you have done is \displaystyle \begin{align*} \frac{f(x + h) - f(h)}{h} \end{align*}, not \displaystyle \begin{align*} \frac{f(x + h) - f(x)}{h} \end{align*}...

6. ## Re: Rate of change?

Originally Posted by Prove It
What you have done is \displaystyle \begin{align*} \frac{f(x + h) - f(h)}{h} \end{align*}, not \displaystyle \begin{align*} \frac{f(x + h) - f(x)}{h} \end{align*}...
yeah, thanks for pointing that

7. ## Re: Rate of change?

Originally Posted by Amer
yeah, thanks for pointing that
Definitely. Thank you! I think I can figure it out from here

8. ## Re: Rate of change?

Originally Posted by Amer
$f(x) = \frac{x}{x^2+2}$

determine $\frac{f(x+h)-f(x)}{h}$ in the simplest form

$f(x+h) = \frac{x+h}{(x+h)^2+2}$

and $f(x) = \frac{x}{x^2+2}$
so

$\frac{f(x+h)-f(x)}{h} = \dfrac{\dfrac{x+h}{(x+h)^2+2} - \dfrac{x}{x^2+2}}{h}$

try to cancel the h in the denominator
Okay.. I'm lost.

Somehow I got to:

(x^3 + hx^2 + 2x +2h) - ((x^2 +hx)^2 +2x)) / h

Am I even on the right track?

9. ## Re: Rate of change?

What happened to the denominators?

$\frac{\frac{x+h}{(x+h)^2+ 2}- \frac{x}{x^2+ 2}}{h}$
Get common denominators:
$\frac{(x+h)(x^2+2)}{h(x^2+2)((x+h)^2+ 2)}- \frac{x((x+h)^2+ 2)}{h(x^2+ 2)((x+h)^2+ 2)}$
$\frac{x^3+ hx^2+ 2x+ 2h- (x^3+ 2hx^2+ (h^2+2)x}{h(x^2+ 2)((x+h)^2+ 2)}$

Reduce the numerator. you can pretty much ignore the denominator. Once you cancel that original "h", the rest of the denominator goes to $(x^2+ 2)^2$.

10. ## Re: Rate of change?

Originally Posted by HallsofIvy
What happened to the denominators?

$\frac{\frac{x+h}{(x+h)^2+ 2}- \frac{x}{x^2+ 2}}{h}$
Get common denominators:
$\frac{(x+h)(x^2+2)}{h(x^2+2)((x+h)^2+ 2)}- \frac{x((x+h)^2+ 2)}{h(x^2+ 2)((x+h)^2+ 2)}$
$\frac{x^3+ hx^2+ 2x+ 2h- (x^3+ 2hx^2+ (h^2+2)x}{h(x^2+ 2)((x+h)^2+ 2)}$

Reduce the numerator. you can pretty much ignore the denominator. Once you cancel that original "h", the rest of the denominator goes to $(x^2+ 2)^2$.

Uhh. Alright. I'm pretty confused on this.. So I ended up with:
$2 - x^2 - hx / x^4 + 2hx^3 + 2h^2 + h^2x^2 + 4x^2 + 4hx +4.$

I think.. I overworked it..