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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

2 Answers 2

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

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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The OP did not ask about the derivative, the OP merely asked about the step before taking the derivative, namely the difference quotient. –  Baby Dragon Jul 4 '13 at 3:01
    
The First Principles Definition of the Derivative is the difference quotient. My hint was just in case the OP was unfamiliar with the formula to calculate it. The difference quotient would give the derivative. –  Sujaan Kunalan Jul 4 '13 at 15:30

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