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.
$$f(x)=x^3 + 2x^2 + 1,\;\;\;c= -2$$
 A: Judging from one of your comments, you’re probably supposed to calculate
$$\begin{align*}
\lim_{x\to-2}\frac{f(x)-f(-2)}{x-(-2)}&=\lim_{x\to-2}\frac{x^3+2x^2+1-\left((-2)^3+2(-2)^2+1\right)}{x+2}\\\\
&=\lim_{x\to-2}\frac{x^3+2x^2}{x+2}\\\\
&=\lim_{x\to-2}\frac{x^2(x+2)}{x+2}\;;
\end{align*}$$
can you finish it from there?
A: 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*}
A: 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 $f'(-2)$?
