Function help

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• Apr 4th 2013, 07:08 PM
Caddym89
Function help
Hi there,

I have a function problem which I'm having trouble solving, and I can't wait until class for the answer.

If f(x) = x^2
and
g(x) = f(x + h) - f(x) all on h
find
g(2)

The answer is apparently 4 + h.

How would I get this answer? I keep getting just h on its own.

thanks
• Apr 4th 2013, 07:13 PM
Gusbob
Re: Function help
Quote:

Originally Posted by Caddym89
Hi there,

I have a function problem which I'm having trouble solving, and I can't wait until class for the answer.

If f(x) = x^2
and
g(x) = f(x + h) - f(x) all on h
find
g(2)

The answer is apparently 4 + h.

How would I get this answer? I keep getting just h on its own.

thanks

I'm confused. If $g(x)=f(x+h)-f(x)$ for all $h$, wouldn't it mean $g(x)=0$ especially when $h=0$?

EDIT: Assuming $g(x)=f(x+h)-f(h)$, we have

$g(2)=f(2+h)-f(h)=(2+h)^2-h^2=4+4h+h^2-h^2=4+4h$
• Apr 4th 2013, 07:25 PM
Caddym89
Re: Function help
Sorry, when I say g(x) = f(x + h) - f(x) all on h, I mean f(x + h) - f(x) is the numerator and h the denominator
• Apr 4th 2013, 07:32 PM
Gusbob
Re: Function help
Oops must have misread that.

Assuming $h\not=0$

$g(2)=\frac{f(2+h)-f(2)}{h}=\frac{(2+h)^2-4}{h}=\frac{4+4h+h^2-4}{h}=\frac{\not{h}(4+h)}{\not{h}}$
• Apr 4th 2013, 07:35 PM
Caddym89
Re: Function help
Thanks a lot Gusbob. I have been looking at this for hours and can't believe it was so simple >_<