use the limit definition of the derivative to find the slope of the tangent line to the curve (f(x)=5x^{2})…

Explanation:

Step1: Find $f(4 + h)$

Given $f(x)=5x^{2}$, substitute $x = 4 + h$: [ \begin{align*} f(4 + h)&=5(4 + h)^{2}\ &=5(16 + 8h+h^{2})\ &=80 + 40h+5h^{2} \end{align*} ]

Step2: Find $f(4)$

Substitute $x = 4$ into $f(x)=5x^{2}$, so $f(4)=5\times4^{2}=5\times16 = 80$.

Step3: Find $f(4 + h)-f(4)$

[ \begin{align*} f(4 + h)-f(4)&=(80 + 40h+5h^{2})-80\ &=40h + 5h^{2} \end{align*} ]

Step4: Find $\frac{f(4 + h)-f(4)}{h}$

[ \begin{align*} \frac{f(4 + h)-f(4)}{h}&=\frac{40h + 5h^{2}}{h}\ &=\frac{h(40 + 5h)}{h}\ &=40+5h \end{align*} ]

Step5: Find $\lim_{h\rightarrow0}\frac{f(4 + h)-f(4)}{h}$

[ \begin{align*} \lim_{h\rightarrow0}\frac{f(4 + h)-f(4)}{h}&=\lim_{h\rightarrow0}(40 + 5h)\ &=40 \end{align*} ] Since $f^{\prime}(4)=\lim_{h\rightarrow0}\frac{f(4 + h)-f(4)}{h}$, $f^{\prime}(4)=40$.

Answer:

$f(4 + h)=80 + 40h+5h^{2}$ $f(4 + h)-f(4)=40h + 5h^{2}$ $\frac{f(4 + h)-f(4)}{h}=40 + 5h$ $\lim_{h\rightarrow0}\frac{f(4 + h)-f(4)}{h}=40$ $f^{\prime}(4)=40$