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.
Hello, Kitty216!
In the numerator, get a common denominator and subtract . . .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.
. . $\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})} $
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.
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