Question

find the indicated quantities for $y = f(x)=4x^{2}$. (a) simplify $\frac{f(1+delta x)-f(1)}{delta x}$. (b) what does the quantity in part (a) approach as $delta x$ approaches 0? (a) $\frac{f(1+delta x)-f(1)}{delta x}=$

Explanation:

Step1: Find \(f(1+\Delta x)\) and \(f(1)\)

Given \(f(x) = 4x^{2}\), then \(f(1+\Delta x)=4(1 + \Delta x)^{2}=4(1 + 2\Delta x+\Delta x^{2})=4 + 8\Delta x+4\Delta x^{2}\), and \(f(1)=4\times1^{2}=4\).

Step2: Substitute into the difference - quotient formula

\(\frac{f(1+\Delta x)-f(1)}{\Delta x}=\frac{(4 + 8\Delta x+4\Delta x^{2})-4}{\Delta x}=\frac{8\Delta x+4\Delta x^{2}}{\Delta x}\).

Step3: Simplify the expression

\(\frac{8\Delta x+4\Delta x^{2}}{\Delta x}=\frac{\Delta x(8 + 4\Delta x)}{\Delta x}=8 + 4\Delta x\).

Step4: Find the limit as \(\Delta x

ightarrow0\)
We want to find \(\lim_{\Delta x
ightarrow0}(8 + 4\Delta x)\). As \(\Delta x\) approaches \(0\), the value of \(4\Delta x\) approaches \(0\). So \(\lim_{\Delta x
ightarrow0}(8 + 4\Delta x)=8\).

Answer:

(A) \(8 + 4\Delta x\)
(B) \(8\)