Question

use the four - step process to find f(x) and then find f(1), f(2), and f(3).
f(x)=x^{2}+4x - 5
f(x)=
f(1)= (type an integer or a simplified fraction.)
f(2)= (type an integer or a simplified fraction.)
f(3)= (type an integer or a simplified fraction.)

Explanation:

Step1: Find $f(x + h)$

$f(x+h)=(x + h)^2+4(x + h)-5=x^{2}+2xh+h^{2}+4x + 4h-5$

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

\[

$$\begin{align*} f(x + h)-f(x)&=(x^{2}+2xh+h^{2}+4x + 4h-5)-(x^{2}+4x-5)\\ &=x^{2}+2xh+h^{2}+4x + 4h-5-x^{2}-4x + 5\\ &=2xh+h^{2}+4h \end{align*}$$

\]

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

\[

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

\]

Step4: Find $\lim_{h

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

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

\]
To find $f^{\prime}(1)$:
Substitute $x = 1$ into $f^{\prime}(x)$: $f^{\prime}(1)=2\times1+4=6$
To find $f^{\prime}(2)$:
Substitute $x = 2$ into $f^{\prime}(x)$: $f^{\prime}(2)=2\times2+4=8$
To find $f^{\prime}(3)$:
Substitute $x = 3$ into $f^{\prime}(x)$: $f^{\prime}(3)=2\times3+4=10$

Answer:

$f^{\prime}(x)=2x + 4$
$f^{\prime}(1)=6$
$f^{\prime}(2)=8$
$f^{\prime}(3)=10$