Question

by determining $f(x)=lim_{h \to 0}\frac{f(x + h)-f(x)}{h}$, find $f(8)$ for the given function
$f(x)=6x^{2}$
$f(8)=square$ (simplify your answer.)

Explanation:

Step1: Find f(x + h)

Given \(f(x)=6x^{2}\), then \(f(x + h)=6(x + h)^{2}=6(x^{2}+2xh+h^{2})=6x^{2}+12xh + 6h^{2}\)

Step2: Substitute into the derivative formula

\[

$$\begin{align*} f^{\prime}(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}\frac{(6x^{2}+12xh + 6h^{2})-6x^{2}}{h}\\ &=\lim_{h ightarrow0}\frac{6x^{2}+12xh + 6h^{2}-6x^{2}}{h}\\ &=\lim_{h ightarrow0}\frac{12xh+6h^{2}}{h}\\ &=\lim_{h ightarrow0}(12x + 6h) \end{align*}$$

\]

Step3: Evaluate the limit

As \(h
ightarrow0\), \(f^{\prime}(x)=12x\)

Step4: Find f'(8)

Substitute \(x = 8\) into \(f^{\prime}(x)\), so \(f^{\prime}(8)=12\times8 = 96\)

Answer:

96