1. ## function help

I plugged f(x) into the problem and got

Im not sure if that's correct. I don't know what to do after this.

2. Hello, Kitty216!

Let: .$\displaystyle f(x)\;=\;\frac{1}{\sqrt{x+1}}$

Find: .$\displaystyle \frac{f(x+h) - f(x)}{h}$ . Simplify as much as possible.

I plugged $\displaystyle f(x)$ into the problem and got: .$\displaystyle \frac{\dfrac{1}{\sqrt{x+h+1}} - \dfrac{1}{\sqrt{x+1}}}{h}$

I'm not sure if that's correct. . . . . It is!
I don't know what to do after this.
In the numerator, get a common denominator and subtract . . .

. . $\displaystyle \frac{1}{\sqrt{x+h+1}}\cdot{\color{blue}\frac{\sqr t{x+1}}{\sqrt{x+1}}} - \frac{1}{\sqrt{x+1}}\cdot{\color{blue}\frac{\sqrt{ x+h+1}}{\sqrt{x+h+1}}} \;\;=\;\;\frac{\sqrt{x+1} - \sqrt{x+h+1}}{\sqrt{x+1}\,\sqrt{x+h+1}}$

So we have: .$\displaystyle \frac{\sqrt{x+1} - \sqrt{x+h+1}}{h\,\sqrt{x+1}\,\sqrt{x+h+1}}$

Multiply top and bottom by: $\displaystyle (\sqrt{x+1} + \sqrt{x+h+1})$

. . $\displaystyle \frac{\sqrt{x+1} - \sqrt{x+h+1}}{h\,\sqrt{x+1}\,\sqrt{x+h+1}} \cdot {\color{blue}\frac{\sqrt{x+1} + \sqrt{x+h+1}}{\sqrt{x+1} + \sqrt{x+h+1}}}$

. . $\displaystyle =\;\frac{(x+1) - (x+h+1)}{h\,\sqrt{x+1}\,\sqrt{x+h+1}\,(\sqrt{x+1} + \sqrt{x+h+1})}$

. . $\displaystyle = \;\frac{-h}{h\,\sqrt{x+1}\,\sqrt{x+h+1}\,(\sqrt{x+1} + \sqrt{x+h+1})}$

. . $\displaystyle =\;\frac{-1}{\sqrt{x+1}\,\sqrt{x+h+1}\,(\sqrt{x+1} + \sqrt{x+h+1})}$

3. That was really helpful =] Is it okay to leave the answer like that? or do I have to rationalize? I think a teacher once told me that I can't have radicals in the denominator.

4. It's ok to leave the answer like that. There is nothing wrong (mathematically) with keeping the radicals in the denominator. The reasoning behind rationalizing fractions is to have some kind of standard form, and to perhaps make the fraction more "simple." But there are many cases where rationalizing the fraction will just make things more complicated, so it may be best to just leave it.