Question

use the four - step process to find f(x) and then find f(-1), f(3), and f(9).
f(x)=2x^{2}+x - 1
f(x)=
f(-1)= (type an integer or a simplified fraction.)
f(3)= (type an integer or a simplified fraction.)
f(9)= (type an integer or a simplified fraction.)

Explanation:

Step1: Find f(x + h)

\[

$$\begin{align*} f(x + h)&=2(x + h)^2+(x + h)-1\\ &=2(x^{2}+2xh+h^{2})+x + h-1\\ &=2x^{2}+4xh+2h^{2}+x + h-1 \end{align*}$$

\]

Step2: Find f(x + h)-f(x)

\[

$$\begin{align*} f(x + h)-f(x)&=(2x^{2}+4xh+2h^{2}+x + h-1)-(2x^{2}+x - 1)\\ &=2x^{2}+4xh+2h^{2}+x + h-1 - 2x^{2}-x + 1\\ &=4xh+2h^{2}+h \end{align*}$$

\]

Step3: Find $\frac{f(x + h)-f(x)}{h}$

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{4xh+2h^{2}+h}{h}\\ &=\frac{h(4x + 2h+1)}{h}\\ &=4x+2h + 1 \end{align*}$$

\]

Step4: Find $\lim_{h

ightarrow0}\frac{f(x + h)-f(x)}{h}$
\[

$$\begin{align*} f^{\prime}(x)&=\lim_{h ightarrow0}(4x+2h + 1)\\ &=4x+1 \end{align*}$$

\]
Now find $f^{\prime}(-1), f^{\prime}(3), f^{\prime}(9)$:

• When $x=-1$, $f^{\prime}(-1)=4\times(-1)+1=-3$
• When $x = 3$, $f^{\prime}(3)=4\times3+1=13$
• When $x = 9$, $f^{\prime}(9)=4\times9+1=37$

Answer:

$f^{\prime}(x)=4x + 1$
$f^{\prime}(-1)=-3$
$f^{\prime}(3)=13$
$f^{\prime}(9)=37$