Here's the question: I am having difficulty finding the difference quotient for: $$y=f(x)=(x-2)^3+4$$
$\endgroup$ 12 Answers
$\begingroup$Notations differ. The difference quotient is $$\frac{f(x+h)-f(x)}{h}.$$ Some people use the symbol $\Delta x$ instead of $h$. So in our case the difference quotient is $$\frac{[ (x+h-2)^3-4] -[(x-2)^3-4]}{h}.$$
You may be expected to simplify this. For the top, the $4$'s cancel, and the top is $(x+h-2)^3-(x-2)^3$.
You may be expected to simplify further. Temporarily, we let $w=x-2$. Then we are looking at $(w+h)^3-w^3$. Expand the first cube. We get $w^3+3w^2h+3wh^2+h^3$. Subtract $w^3$. Now note that each term left has an $h$ in it, so we can cancel with the $h$ at the bottom of the difference quotient. We end up with $$3w^2+3wh+h^2$$ (if $h\ne 0$).
Now for the final answer, replace $w$ by $x-2$, and (perhaps) expand.
$\endgroup$ 4 $\begingroup$Hint: Apply First Principles Definition of the Derivative:
$\lim_{h\to 0 }\frac{f(x+h)-f(x)}{h}\implies\lim_{h\to 0}\frac{(x+h-2)^3+4-((x-2)^3+4)}{h}$
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