Question

a) $lim_{x \to 2}x^{2}+4$

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

find 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 $f(x + h)=2(x + h)^{2}+3(x + h)-1$.

Step2: Expand $f(x + h)$

Expand $2(x + h)^{2}+3(x + h)-1$:
\[

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

\]

Step3: Calculate $f(x + h)-f(x)$

\[

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

\]

Step4: Calculate $\frac{f(x + h)-f(x)}{h}$

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

Step5: Find the limit as $h

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

Step6: Evaluate $f^{\prime}(x)$ at $x = 3$

Substitute $x = 3$ into $f^{\prime}(x)$: $f^{\prime}(3)=4\times3+3=12 + 3=15$.

Answer:

15