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) = limlimits_{h \to 0} \frac{f(a + h) - f(a)}{h} ).
b. find the equation for the tangent line to ( y = w(x) ) at ( x = 2 ).

Explanation:

Part A

Step1: Identify \(a\), \(f(a+h)\) and \(f(a)\)

Here, \(a = 2\) and \(w(x)=5x^{2}+3x + 3\). So, \(w(2+h)=5(2 + h)^{2}+3(2 + h)+3\) and \(w(2)=5(2)^{2}+3(2)+3\).
First, expand \(w(2 + h)\):
\[

$$\begin{align*} w(2+h)&=5(4 + 4h+h^{2})+6 + 3h+3\\ &=20+20h + 5h^{2}+6 + 3h+3\\ &=5h^{2}+23h + 29 \end{align*}$$

\]
And \(w(2)=5\times4+6 + 3=20 + 6+3 = 29\).

Step2: Substitute into the limit formula

The limit definition is \(w^{\prime}(2)=\lim_{h
ightarrow0}\frac{w(2 + h)-w(2)}{h}\). Substitute \(w(2 + h)=5h^{2}+23h + 29\) and \(w(2) = 29\) into it:
\[

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

\]

Step3: Simplify the expression

Factor out \(h\) from the numerator: \(\frac{5h^{2}+23h}{h}=\frac{h(5h + 23)}{h}\). Since \(h
eq0\) (we are taking the limit as \(h
ightarrow0\), not evaluating at \(h = 0\)), we can cancel \(h\):
\[
w^{\prime}(2)=\lim_{h
ightarrow0}(5h + 23)
\]

Step4: Evaluate the limit

As \(h
ightarrow0\), substitute \(h = 0\) into \(5h+23\): \(5(0)+23=23\).

Step1: Find the point of tangency

We know that at \(x = 2\), \(y=w(2)\). From part A, \(w(2)=29\), so the point is \((2,29)\).

Step2: Recall the slope of the tangent line

The slope \(m\) of the tangent line to \(y = w(x)\) at \(x = 2\) is \(w^{\prime}(2)\). From part A, \(m = 23\).

Step3: Use the point - slope form of a line

The point - slope form of a line is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})=(2,29)\) and \(m = 23\). Substitute these values:
\[

$$\begin{align*} y-29&=23(x - 2) \end{align*}$$

\]

Step4: Simplify to slope - intercept form (optional, but common)

Expand the right - hand side: \(y-29=23x-46\). Then add 29 to both sides: \(y=23x-46 + 29=23x-17\).

Answer:

\(w^{\prime}(2)=23\)

Part B