Find the difference quotient for the given function.
f(x) = sqrt{x} + 1
The difference quotient is defined as $\displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{h}$ (When you study calculus, taking the limit as h approaches zero of the difference quotient will yield the derivative).
Since $\displaystyle f\left(x\right)=\sqrt{x}+1$, we see that the difference quotient is $\displaystyle \frac{\sqrt{x+h}+1-\left(\sqrt{x}+1\right)}{h}=\frac{\sqrt{x+h}-\sqrt{x}}{h}$
We can try to manipulate this a bit.
Multiplying both numerator and denominator by the conjugate $\displaystyle \sqrt{x+h}+\sqrt{x}$, we get $\displaystyle \frac{x+h-x}{h\left(\sqrt{x+h}+\sqrt{x}\right)}$ which yields $\displaystyle \color{red}\boxed{\frac{1}{\sqrt{x+h}+\sqrt{x}}}$
Does this make sense?
Good Question!
At this stage, I believe it would be fine to leave it as $\displaystyle \frac{\sqrt{x+h}-\sqrt{x}}{h}$
But as you go further on in math, it will be a little clearer why you would get it into the final form I have it in.
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(As an aside)
When you're in calculus and you need to evaluate $\displaystyle \lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$, it is best to give it in the final form I have, for you can then easily evaluate the limit [the reason why we manipulate the original limit by multiplying it by the conjugate of the numerator is to avoid dividing by zero].
This probably will make more sense when you get closer to studying calculus.