use the four - step process to find f(x) and then find f(1), f(2), and f(3).\nf(x)=x^{2}+9x - 4\nf(x)=□

Explanation:

Step1: Find $f(x + h)$

$f(x+h)=(x + h)^2+9(x + h)-4=x^{2}+2xh+h^{2}+9x + 9h-4$

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

[ \begin{align*} f(x + h)-f(x)&=(x^{2}+2xh+h^{2}+9x + 9h-4)-(x^{2}+9x-4)\ &=x^{2}+2xh+h^{2}+9x + 9h-4-x^{2}-9x + 4\ &=2xh+h^{2}+9h \end{align*} ]

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

$\frac{f(x + h)-f(x)}{h}=\frac{2xh+h^{2}+9h}{h}=2x + h+9$

Step4: Find $\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$

$f^{\prime}(x)=\lim_{h\rightarrow0}(2x + h+9)=2x+9$

To find $f^{\prime}(1)$: Substitute $x = 1$ into $f^{\prime}(x)$: $f^{\prime}(1)=2\times1+9=11$ To find $f^{\prime}(2)$: Substitute $x = 2$ into $f^{\prime}(x)$: $f^{\prime}(2)=2\times2+9=13$ To find $f^{\prime}(3)$: Substitute $x = 3$ into $f^{\prime}(x)$: $f^{\prime}(3)=2\times3+9=15$

Answer:

$f^{\prime}(x)=2x + 9$, $f^{\prime}(1)=11$, $f^{\prime}(2)=13$, $f^{\prime}(3)=15$