# I need to find the difference quotient for a function

Here's the question: I am having difficulty finding the difference quotient for: $$y=f(x)=(x-2)^3+4$$

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Are you familiar with the definition of the difference quotient? You should show us the work you have so far so that we can better help you. –  Spencer Jul 4 '13 at 2:00

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.

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Is this seem correct for the final answer: 3x^2-12x+12+3xh-6h+h^2 –  sawblade Jul 4 '13 at 3:43
Yes, it is correct. To find the derivative from first principles, we let $h\to 0$, get $3x^2-12x+12$. Myself, I would not simplify so far, and would leave the difference quotient as $3(x-2)^2+3h(x-2)+h^2$. But I do not know what kind of simplification is expected of you. –  André Nicolas Jul 4 '13 at 4:00
Alright! Thank you for all of your help. –  sawblade Jul 4 '13 at 4:08
You are welcome. This first part of introduction to derivatives gives your algebra a workout. Alternatively (check it) I could have used the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$, with $a=x+h-2$ and $b=x-2$. It is "simpler" but somewhat more sophisticated than crude expansion, –  André Nicolas Jul 4 '13 at 4:13
$\lim_{h\to 0 }\frac{f(x+h)-f(x)}{h}\implies\lim_{h\to 0}\frac{(x+h-2)^3+4-((x-2)^3+4)}{h}$