Question

1. use the limit definition of derivative to find $f(x)$ for $f(x)=2x^{2}+3x - 1$.

b) give the equation of the line tangent to $y = 2x^{2}+3x - 1$ at $x = 3$.

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative is $f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=2x^{2}+3x - 1$, then $f(x + h)=2(x + h)^{2}+3(x + h)-1$.
First, expand $f(x + h)$:
\[

$$\begin{align*} f(x + h)&=2(x^{2}+2xh+h^{2})+3x + 3h-1\\ &=2x^{2}+4xh+2h^{2}+3x + 3h-1 \end{align*}$$

\]

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

\[

$$\begin{align*} f(x + h)-f(x)&=(2x^{2}+4xh+2h^{2}+3x + 3h-1)-(2x^{2}+3x - 1)\\ &=2x^{2}+4xh+2h^{2}+3x + 3h-1 - 2x^{2}-3x + 1\\ &=4xh+2h^{2}+3h \end{align*}$$

\]

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

\[
\frac{f(x + h)-f(x)}{h}=\frac{4xh+2h^{2}+3h}{h}=4x + 2h+3
\]

Step4: Find the limit as $h

ightarrow0$
\[
f^{\prime}(x)=\lim_{h
ightarrow0}(4x + 2h+3)=4x+3
\]

Step5: Find the slope of the tangent line at $x = 3$

Substitute $x = 3$ into $f^{\prime}(x)$. So $m=f^{\prime}(3)=4\times3+3=15$.

Step6: Find the point on the curve at $x = 3$

Substitute $x = 3$ into $y = f(x)=2x^{2}+3x - 1$. Then $y=2\times3^{2}+3\times3-1=2\times9 + 9-1=18 + 9-1=26$. The point is $(3,26)$.

Step7: Use the point - slope form to find the equation of the tangent line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(3,26)$ and $m = 15$.
\[
y-26=15(x - 3)
\]
\[
y-26=15x-45
\]
\[
y=15x-19
\]

Answer:

a) $f^{\prime}(x)=4x + 3$
b) $y=15x-19$