I have the problem Find the difference quotient $\frac{f(2 + h) - f(2)}{h}$ for $f(x) = \frac{1}{x^2}$.
The answer they gave is $\frac{-(4 + h)}{4(2 + h)^2}$
So far I've done:
$$\frac{[1/(2 + h)^2 - 1/4] }{ h}$$
$$\frac{[-4/4(2 + h)^2 - (2 + h)^2/4(2 + h)^2]}{ h} $$
so the $4(2 + h)^2$ is the solution's denominator based on that, which is right,...
So then that leaves $\frac{[-4 - (2 + h)^2]}{ h}$ or $\frac{[-4 - (4 + h^2)]}{ h}$ for finding the top half of the answer. Which is where I'm having a problem.
for $f(x)=\frac{1}{x^2}$ $$\frac{f(2+h)-f(2)}{h}=\frac{\frac{1}{(2+h)^2}-\frac{1}{4}}{h}=\frac{\frac{4-(2+h)^2}{4(2+h)^2}}{h}$$ $$=\frac{4-(2+h)^2}{4h(2+h)^2}=\frac{4-(4+4h+h^2)}{4h(2+h)^2}$$ $$=\frac{-4h-h^2}{4h(2+h)^2}=\frac{-h(4+h)}{4h(2+h)^2}=\frac{-(4+h)}{4(2+h)^2}$$ Which is the answer as required.