QUESTION IMAGE
Question
use the four - step process to find r(x) and then find r(1), r(2), and r(3).
r(x)=6 - 4x^{2}
r(x)=
(type an integer or a simplified fraction.)
r(1)=
(type an integer or a simplified fraction.)
r(2)=
(type an integer or a simplified fraction.)
r(3)=
(type an integer or a simplified fraction.)
Step1: Recall the four - step process for finding the derivative
The four - step process for finding the derivative of a function $y = r(x)$ is based on the definition of the derivative $r^{\prime}(x)=\lim_{h
ightarrow0}\frac{r(x + h)-r(x)}{h}$. First, find $r(x + h)$:
Given $r(x)=6-4x^{2}$, then $r(x + h)=6-4(x + h)^{2}=6-4(x^{2}+2xh+h^{2})=6-4x^{2}-8xh - 4h^{2}$.
Step2: Calculate $r(x + h)-r(x)$
$r(x + h)-r(x)=(6-4x^{2}-8xh - 4h^{2})-(6 - 4x^{2})=-8xh-4h^{2}$.
Step3: Calculate $\frac{r(x + h)-r(x)}{h}$
$\frac{r(x + h)-r(x)}{h}=\frac{-8xh - 4h^{2}}{h}=-8x-4h$.
Step4: Find the limit as $h
ightarrow0$
$r^{\prime}(x)=\lim_{h
ightarrow0}\frac{r(x + h)-r(x)}{h}=\lim_{h
ightarrow0}(-8x - 4h)=-8x$.
Now, find $r^{\prime}(1),r^{\prime}(2),r^{\prime}(3)$:
- For $x = 1$, $r^{\prime}(1)=-8\times1=-8$.
- For $x = 2$, $r^{\prime}(2)=-8\times2=-16$.
- For $x = 3$, $r^{\prime}(3)=-8\times3=-24$.
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$r^{\prime}(x)=-8x$
$r^{\prime}(1)=-8$
$r^{\prime}(2)=-16$
$r^{\prime}(3)=-24$