Question

question let (f(x)=-3x^{2}-6x - 13). find (f(3)) using the definition of a derivative. provide your answer below: (f(3)=square)

Explanation:

Step1: Recall derivative definition

The definition of the derivative of a function $y = f(x)$ at a point $x=a$ is $f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here $a = 3$ and $f(x)=-3x^{2}-6x - 13$. First, find $f(3 + h)$ and $f(3)$.
\[

$$\begin{align*} f(3 + h)&=-3(3 + h)^{2}-6(3 + h)-13\\ &=-3(9 + 6h+h^{2})-18-6h - 13\\ &=-27-18h-3h^{2}-18 - 6h-13\\ &=-3h^{2}-24h - 58 \end{align*}$$

\]
\[
f(3)=-3\times3^{2}-6\times3-13=-3\times9 - 18-13=-27-18 - 13=-58
\]

Step2: Substitute into derivative - formula

\[

$$\begin{align*} f^{\prime}(3)&=\lim_{h ightarrow0}\frac{f(3 + h)-f(3)}{h}\\ &=\lim_{h ightarrow0}\frac{-3h^{2}-24h - 58-(-58)}{h}\\ &=\lim_{h ightarrow0}\frac{-3h^{2}-24h - 58 + 58}{h}\\ &=\lim_{h ightarrow0}\frac{-3h^{2}-24h}{h} \end{align*}$$

\]

Step3: Simplify the expression

\[

$$\begin{align*} \lim_{h ightarrow0}\frac{-3h^{2}-24h}{h}&=\lim_{h ightarrow0}\frac{h(-3h - 24)}{h}\\ &=\lim_{h ightarrow0}(-3h - 24) \end{align*}$$

\]

Step4: Evaluate the limit

As $h
ightarrow0$, we substitute $h = 0$ into $-3h - 24$. So $f^{\prime}(3)=-24$.

Answer:

$-24$