Question

use the limit definition of the derivative to find the slope of the tangent line to the curve f(x) = 7x² + 6x + 2 at x = 3

Explanation:

Step1: Recall the limit definition of the derivative

The limit definition of the derivative at a point \(x = a\) is \(f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}\). Here, \(a = 3\) and \(f(x)=7x^{2}+6x + 2\). First, find \(f(3+h)\) and \(f(3)\).

\(f(3+h)=7(3 + h)^{2}+6(3 + h)+2\)
Expand \((3 + h)^{2}=9 + 6h+h^{2}\), so \(f(3+h)=7(9 + 6h+h^{2})+18 + 6h+2=63+42h + 7h^{2}+18+6h + 2=7h^{2}+48h + 83\)

\(f(3)=7(3)^{2}+6(3)+2=7\times9 + 18+2=63 + 18+2 = 83\)

Step2: Substitute into the limit formula

\(f^{\prime}(3)=\lim_{h
ightarrow0}\frac{f(3 + h)-f(3)}{h}=\lim_{h
ightarrow0}\frac{(7h^{2}+48h + 83)-83}{h}=\lim_{h
ightarrow0}\frac{7h^{2}+48h}{h}\)

Step3: Simplify the expression

Factor out \(h\) from the numerator: \(\lim_{h
ightarrow0}\frac{h(7h + 48)}{h}\). Since \(h
eq0\) (we are taking the limit as \(h
ightarrow0\), not evaluating at \(h = 0\)), we can cancel \(h\) from numerator and denominator: \(\lim_{h
ightarrow0}(7h + 48)\)

Step4: Evaluate the limit

As \(h
ightarrow0\), substitute \(h = 0\) into \(7h+48\): \(7(0)+48=48\)

Answer:

\(48\)