Question

hw 7 - definition of the derivative section 2.2: problem 3 (1 point)
let $f(x)=1 - 6x^{2}$. 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 $f(x)=1 - 6x^{2}$, then $f(x + h)=1-6(x + h)^{2}=1-6(x^{2}+2xh+h^{2})=1 - 6x^{2}-12xh - 6h^{2}$.

Step2: Calculate the difference - quotient

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{(1 - 6x^{2}-12xh - 6h^{2})-(1 - 6x^{2})}{h}\\ &=\frac{1 - 6x^{2}-12xh - 6h^{2}-1 + 6x^{2}}{h}\\ &=\frac{-12xh - 6h^{2}}{h}\\ &=-12x-6h \end{align*}$$

\]
Comparing with $Ah + Bx + C$, we have $A=-6$, $B=-12$, $C = 0$.

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

\[

$$\begin{align*} f'(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}(-12x-6h)\\ &=-12x \end{align*}$$

\]

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

When $x = 1$, $f'(1)=-12\times1=-12$.
When $x = 2$, $f'(2)=-12\times2=-24$.
When $x = 3$, $f'(3)=-12\times3=-36$.

Answer:

$A=-6$, $B=-12$, $C = 0$, $f'(x)=-12x$, $f'(1)=-12$, $f'(2)=-24$, $f'(3)=-36$