# Need help using the alternative formula to find the derivative at $x = c$

Don't know if I'm simplifying wrong or if it is a simple mistake but I cannot get the answer of 4. Please help and include exact step by step details. Thank you.

$$f(x)=x^3 + 2x^2 + 1,\;\;\;c= -2$$

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What do you mean by alternative formula? Could you be more specific? – amWhy Jul 21 '13 at 1:44
Better yet, could you please edit to include what you've tried? That will help us to clear up any confusion. – amWhy Jul 21 '13 at 1:52

\begin{align*} \lim_{x\to-2}\frac{f(x)-f(-2)}{x-(-2)}&=\lim_{x\to-2}\frac{x^3+2x^2+1-\left((-2)^3+2(-2)^2+1\right)}{x+2}\\\\ &=\lim_{x\to-2}\frac{x^3+2x^2}{x+2}\\\\ &=\lim_{x\to-2}\frac{x^2(x+2)}{x+2}\;; \end{align*}

can you finish it from there?

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Wow thank you so much. I completely see where I was messing it up. Thank you so much for the help, it is very greatly appreciated Brian. Thank you again. – Nick Jul 21 '13 at 2:01
@Nick: You’re very welcome. – Brian M. Scott Jul 21 '13 at 2:02

Using the alternative definition of the derivative, given what you posted in a comment, we'll start with the approximation of the derivative given by $\lim_{x \to -2}\dfrac{f(x) - f(c)}{x - c}$

\begin{align*}\require{cancel} \lim_{x \to -2} \dfrac{f(x) - f(c)}{x - c} &=\lim_{x\to -2} \dfrac{x^3 + 2x^2 + 1 - ((-2)^3 + 2(-2)^2 + 1)}{x - (-2)}\\ &=\lim_{x \to -2} \dfrac{x^3 + 2x^2 + 1 -(-8 + 2\cdot 4 + 1)}{x + 2}\\ &=\lim_{x \to -2} \dfrac{x^3 + 2x^2 + 1 - (1)}{x + 2}\\ &= \lim_{x \to -2} \dfrac{x^2({x + 2)}}{{x+2}}\\ &= \lim_{x \to -2} \dfrac{x^2\cancel{(x + 2)}}{\cancel{x+2}} \\ &= \lim_{x \to -2} x^2 = (-2)^2 = 4 \end{align*}

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Thanks, for the editing @dfeuer! Looks much "prettier" to look at now :-) – amWhy Jul 21 '13 at 3:42
You're welcome. Note that you need to \require{cancel} before you can use those nifty cancellations. – dfeuer Jul 21 '13 at 3:58

If $f(x) = x^3 + 2x^2 + 1$, then $f'(x) = 3x^2 + 4x + 0$ from the power rule. Substituting $-2$ for $x$ we have that $f'(-2) = 3(-2)^2 + 4(-2) = 12 - 8 = 4$. On a side note, what value are you getting for $f'(-2)$?

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I need it using the Alternative Form of the derivative, please and thank you. – Nick Jul 21 '13 at 1:48
I am getting zero when using the alternative form – Nick Jul 21 '13 at 1:49
Would you mind telling us what the "alternative form" is? I don't know that this is standard nomenclature. – Cameron Williams Jul 21 '13 at 1:49
f(x)-f(c)/x-c is what it shows in my calculus book – Nick Jul 21 '13 at 1:53
That's not a derivative. That is the formula for the slope of a secant line - a related, but different concept. – Cameron Williams Jul 21 '13 at 1:55