Question

hw 7 - definition of the derivative section 2.2: problem 6 (1 point)
let $f(x)=6 + 6x-4x^{2}$. 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 $f(x)=6 + 6x-4x^{2}$, then $f(x + h)=6+6(x + h)-4(x + h)^{2}$.
Expand $f(x + h)$:
\[

$$\begin{align*} f(x + h)&=6+6x+6h-4(x^{2}+2xh + h^{2})\\ &=6+6x+6h-4x^{2}-8xh - 4h^{2} \end{align*}$$

\]

Step2: Calculate $\frac{f(x + h)-f(x)}{h}$

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{(6 + 6x+6h-4x^{2}-8xh - 4h^{2})-(6 + 6x-4x^{2})}{h}\\ &=\frac{6 + 6x+6h-4x^{2}-8xh - 4h^{2}-6 - 6x + 4x^{2}}{h}\\ &=\frac{6h-8xh - 4h^{2}}{h}\\ &=6-8x - 4h \end{align*}$$

\]
Comparing with $Ah + Bx+C$, we have $A=-4$, $B=-8$, $C = 6$.

Step3: Find $f'(x)$

\[

$$\begin{align*} f'(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}(6-8x - 4h)\\ &=6-8x \end{align*}$$

\]

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

For $x = 1$, $f'(1)=6-8\times1=-2$.
For $x = 2$, $f'(2)=6-8\times2=6 - 16=-10$.
For $x = 3$, $f'(3)=6-8\times3=6 - 24=-18$.

Answer:

$A=-4$, $B=-8$, $C = 6$, $f'(x)=6 - 8x$, $f'(1)=-2$, $f'(2)=-10$, $f'(3)=-18$