Simplify the difference quotient $\frac{f(x+h)-f(x)}{h}$ where
a) $f(x)=2x+3,$
b) $f(x)=\frac{1}{x+1},$
c) $f(x)=x^2.$
I believe that if anyone can help me out with the first one, the other two might come clearer to me. But I started out this problem by plugging in $f(x)$ and got: $\dfrac{(2x+3)+f(h)-(2x+3)}{h}$ I have no idea what to do after this. Because the $2x+3$'s can cancel out and leave me with just $\frac{f(h)}{h}$ but that doesn't make sense to me. Please help.
EDIT: The first one is solved and now for the second one, this is what I got: $\dfrac{\frac{1}{(x+h)+1}-\frac{1}{x+1}}{h}$ Now, to subtract fractions the denominator has to be the same and they are the exact same except for the first fraction has an $h$ while the other one does not. How would I go about this?
For the third problem, this is all of my work: $\begin{align*}\dfrac{(x+h)^2-x^2}{h}&= \dfrac{x^2+2hx+h^2-x^2}{h}\\ &= \dfrac{2hx+h^2}{h}\\ &= \dfrac{h(2x+h)}{h}\\ &= 2x+h\end{align*}$ Is this all correct?