Question

if f(x)=√(x+4) find and simplify f(2+h)-f(2)/h

Answer 1

I already have it down to (√(6+h)-√(6))/h, but the answer given to me is 1/(√6+h)+√(6)) and I can't figure out why.

Answer 2

\[\Large\rm f(x)=\sqrt{x+4}\]\[\Large\rm \frac{f(2+h)-f(2)}{h}\quad=\quad \frac{\sqrt{x+6}-\sqrt{6}}{h}\]We'll multiply the top and bottom by the conjugate of the numerator,\[\Large\rm \frac{\sqrt{x+6}-\sqrt{6}}{h}\color{royalblue}{\left(\frac{\sqrt{x+6}+\sqrt{6}}{\sqrt{x+6}+\sqrt{6}}\right)}\]

Answer 3

Do you understand what that will do to our numerator?

Answer 4

Oh okay! So, when foiling the numerator, it'll cancel out resulting in 1?

Answer 5

I just wrote it down. Some of the numerator will cancel out with itself when factored and the rest will cancel out with the bottom portion, is that right?

Answer 6

When you multiply conjugates, you're left with the difference of squares.
If you want to FOIL it though, that's fine, same result.

You should get the first term squared, minus the second term squared.
So the roots fall off.
And yah a bunch of stuff cancels out,\[\Large\rm \frac{h+6-6}{h\left(\sqrt{h+6}+\sqrt 6\right)}=\frac{\cancel h}{\cancel h\left(\sqrt{h+6}+\sqrt 6\right)}\]Yah looks like a 1 left on top :)
Bahhhh I wrote x's under the root earlier, sorry bout that.

Answer 7

It's all good, thanks for the help!