# Functions (Limits)

• Jul 28th 2009, 03:40 PM
drkidd22
Functions (Limits)
So...

I'm a little confused as of what I got to do here but this is what I've done so far.

$\displaystyle f(x)$ is given. Find $\displaystyle f(x+\Delta x)$.

$\displaystyle f(x) = x^2+1$

$\displaystyle (x+\Delta x)^2 + 1$

$\displaystyle (x+\Delta x)(x+\Delta x)+1$

$\displaystyle = x^2+2x\Delta x+\Delta x^2+1$

So...That it I'm done?
• Jul 28th 2009, 04:13 PM
pickslides
= $\displaystyle (1+\Delta) x^2+2x\Delta x+1$

• Jul 28th 2009, 04:23 PM
drkidd22
nope, no limit asked, just said open up parentheses that why I left it like that
• Jul 28th 2009, 04:27 PM
ANDS!
Quote:

Originally Posted by drkidd22
So...

I'm a little confused as of what I got to do here but this is what I've done so far.

$\displaystyle f(x)$ is given. Find $\displaystyle f(x+\Delta x)$.

$\displaystyle f(x) = x^2+1$

$\displaystyle (x+\Delta x)^2 + 1$

$\displaystyle (x+\Delta x)(x+\Delta x)+1$

$\displaystyle = x^2+2x\Delta x+\Delta x^2+1$

So...That it I'm done?

Are you meant to find the difference quotient?
• Jul 28th 2009, 04:33 PM
drkidd22
Just says simplify completly
• Jul 28th 2009, 04:39 PM
drkidd22
I'm stuck on this simplifying this one.

$\displaystyle f(x)=\sqrt{x^2-1}$

$\displaystyle =\sqrt{(x+\Delta x)^2-1}$

$\displaystyle =\sqrt{x^2+2x\Delta x+\Delta x^2-1}$
• Jul 28th 2009, 04:48 PM
skeeter
Quote:

Originally Posted by drkidd22
I'm stuck on this simplifying this one.

$\displaystyle f(x)=\sqrt{x^2-1}$

$\displaystyle =\sqrt{(x+\Delta x)^2-1}$

$\displaystyle =\sqrt{x^2+2x\Delta x+\Delta x^2-1}$

you're done.