Question

1. let $w(x)=5x^{2}+3x + 3$.

a. use the limit definition of the derivative to find $w(2)$.
use the form: $f(a)=lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$.
b. find the equation for the tangent line to $y = w(x)$ at $x = 2$.

Explanation:

Step1: Find \(w(2 + h)\) and \(w(2)\)

First, find \(w(2+h)=5(2 + h)^{2}+3(2 + h)+3\). Expand \((2 + h)^{2}=4 + 4h+h^{2}\), so \(w(2 + h)=5(4 + 4h+h^{2})+6 + 3h+3=20+20h + 5h^{2}+6 + 3h+3=5h^{2}+23h + 29\). And \(w(2)=5\times2^{2}+3\times2 + 3=20 + 6+3=29\).

Step2: Apply the limit - definition of the derivative

\[

$$\begin{align*} w^{\prime}(2)&=\lim_{h ightarrow0}\frac{w(2 + h)-w(2)}{h}\\ &=\lim_{h ightarrow0}\frac{5h^{2}+23h + 29 - 29}{h}\\ &=\lim_{h ightarrow0}\frac{5h^{2}+23h}{h}\\ &=\lim_{h ightarrow0}(5h + 23) \end{align*}$$

\]
As \(h
ightarrow0\), \(w^{\prime}(2)=23\).

Step3: Find the point on the curve

When \(x = 2\), \(y=w(2)=29\), so the point \((x_1,y_1)=(2,29)\).

Step4: Find the equation of the tangent line

The equation of a line is \(y - y_1=m(x - x_1)\), where \(m = w^{\prime}(2)=23\), \(x_1 = 2\) and \(y_1=29\). So \(y-29=23(x - 2)\), which simplifies to \(y-29=23x-46\), or \(y = 23x-17\).

Answer:

A. \(w^{\prime}(2)=23\)
B. \(y = 23x-17\)