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} ).

Explanation:

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

Here, \(a = 2\), \(f(x)=w(x)=5x^{2}+3x + 3\). First, find \(w(2+h)\) and \(w(2)\).

\(w(2 + h)=5(2 + h)^{2}+3(2 + h)+3\)
Expand \((2 + h)^{2}=4 + 4h+h^{2}\), so:
\(w(2 + h)=5(4 + 4h+h^{2})+6 + 3h+3=20+20h + 5h^{2}+6 + 3h+3=5h^{2}+23h + 29\)

\(w(2)=5(2)^{2}+3(2)+3=5\times4 + 6 + 3=20 + 6+3 = 29\)

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 \(w(2)\) into the formula:
\(\lim_{h
ightarrow0}\frac{(5h^{2}+23h + 29)-29}{h}=\lim_{h
ightarrow0}\frac{5h^{2}+23h}{h}\)

Step3: Simplify the expression

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

Step4: Evaluate the limit

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

Answer:

\(23\)