use the four-step definition of the derivative to find $f(x)$ if $f(x) = -3x^2 - 2x + 17$.
$f(x + h) = \square$
$f(x + h) - f(x) = \square$
$\frac{f(x + h) - f(x)}{h} = \square$
find $f(x)$ by determining $\lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} = \square$
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Question

Accepted Answer

# Explanation:
## Step1: Find $ f(x + h) $
Substitute $ x + h $ into $ f(x) = -3x^2 - 2x + 17 $.
$$
\begin{align*}
f(x + h)&=-3(x + h)^2 - 2(x + h) + 17\
&=-3(x^2 + 2xh + h^2) - 2x - 2h + 17\
&=-3x^2 - 6xh - 3h^2 - 2x - 2h + 17
\end{align*}
$$

## Step2: Find $ f(x + h) - f(x) $
Subtract $ f(x) $ from $ f(x + h) $.
$$
\begin{align*}
f(x + h) - f(x)&=(-3x^2 - 6xh - 3h^2 - 2x - 2h + 17) - (-3x^2 - 2x + 17)\
&=-3x^2 - 6xh - 3h^2 - 2x - 2h + 17 + 3x^2 + 2x - 17\
&=-6xh - 3h^2 - 2h
\end{align*}
$$

## Step3: Find $ \frac{f(x + h) - f(x)}{h} $
Divide $ f(x + h) - f(x) $ by $ h $ ( $ h
eq 0 $ ).
$$
\begin{align*}
\frac{f(x + h) - f(x)}{h}&=\frac{-6xh - 3h^2 - 2h}{h}\
&=\frac{h(-6x - 3h - 2)}{h}\
&=-6x - 3h - 2
\end{align*}
$$

## Step4: Find $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $
Take the limit as $ h \to 0 $.
$$
\begin{align*}
\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}&=\lim_{h \to 0} (-6x - 3h - 2)\
&=-6x - 3(0) - 2\
&=-6x - 2
\end{align*}
$$

# Answers:
- $ f(x + h) = -3x^2 - 6xh - 3h^2 - 2x - 2h + 17 $
- $ f(x + h) - f(x) = -6xh - 3h^2 - 2h $
- $ \frac{f(x + h) - f(x)}{h} = -6x - 3h - 2 $
- $ f'(x) = -6x - 2 $

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QUESTION IMAGE

Question

Explanation:

Step1: Find \( f(x + h) \)

Step2: Find \( f(x + h) - f(x) \)

Step3: Find \( \frac{f(x + h) - f(x)}{h} \)

Step4: Find \( \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \)

Answer: