Question

hw 7 - definition of the derivative section 2.2: problem 2 (1 point)
let $f(x)=4 - 8x$. if $h
eq0$, then the difference quotient can be simplified as
$\frac{f(x + h)-f(x)}{h}=ah + bx + c$,
where $a$, $b$, and $c$ are constants. (note: its possible for one or more of these constants to be 0.) find the constants.
$a=square$, $b=square$, and $c=square$.
use the simplified expression to find $f(x)=lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=square$.
finally, find each of the following.
$f(1)=square$, $f(2)=square$, and $f(3)=square$.

Explanation:

Step1: Find $f(x + h)$

Given $f(x)=4 - 8x$, then $f(x + h)=4-8(x + h)=4-8x-8h$.

Step2: Calculate the difference - quotient

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{(4-8x - 8h)-(4 - 8x)}{h}\\ &=\frac{4-8x - 8h-4 + 8x}{h}\\ &=\frac{-8h}{h}\\ &=-8 \end{align*}$$

\]
Since $\frac{f(x + h)-f(x)}{h}=Ah + Bx + C=-8$, comparing coefficients, we have $A = 0$, $B = 0$, $C=-8$.

Step3: Find the derivative $f'(x)$

\[
f'(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}
\]
Since $\frac{f(x + h)-f(x)}{h}=-8$ for $h
eq0$, then $f'(x)=\lim_{h
ightarrow0}(-8)=-8$.

Step4: Evaluate $f'(1)$, $f'(2)$ and $f'(3)$

Since $f'(x)=-8$ is a constant function, then $f'(1)=-8$, $f'(2)=-8$, $f'(3)=-8$.

Answer:

$A = 0$
$B = 0$
$C=-8$
$f'(x)=-8$
$f'(1)=-8$
$f'(2)=-8$
$f'(3)=-8$