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 \( w(2 + h) \):
\( w(2 + h)=5(4 + 4h+h^{2})+6 + 3h+3=20 + 20h+5h^{2}+6 + 3h+3=5h^{2}+23h + 29 \)
And \( w(2)=5\times4+6 + 3=20 + 6+3 = 29 \)

Step2: Substitute into the limit formula

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

Step3: Simplify the expression

Factor out \( h \) from the numerator:
\( \lim_{h
ightarrow0}\frac{h(5h + 23)}{h} \)
Cancel out \( h \) (since \( h
eq0 \) when taking the limit):
\( \lim_{h
ightarrow0}(5h + 23) \)

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 \( x = 2 \). Find \( y=w(2) \). From part A, \( w(2)=29 \). So the point is \( (2,29) \).

Step2: Find the slope of the tangent line

The slope of the tangent line at \( x = 2 \) is \( w^{\prime}(2) \), which we found in part A as \( 23 \).

Step3: Use the point - slope form of a line

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

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

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

Answer:

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

Part B