Question

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 $\lim\limits_{h \to 0} \frac{f(x + h) - f(x)}{h} = \square$
question help: \boxed{video}

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 \( h \to 0 \).
\[

$$\begin{align*} \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}&=\lim_{h \to 0} (-6x - 3h - 2)\\ &=-6x - 3(0) - 2\\ &=-6x - 2 \end{align*}$$

\]

Answer:

s:

• \( f(x + h) = -3x^2 - 6xh - 3h^2 - 2x - 2h + 17 \)
• \( f(x + h) - f(x) = -6xh - 3h^2 - 2h \)
• \( \frac{f(x + h) - f(x)}{h} = -6x - 3h - 2 \)
• \( f'(x) = -6x - 2 \)