Question

use the limit definition of the derivative to find the slope of the tangent line to the curve (f(x)=5x^{2}) at (x = 4). evaluate each of the following and enter your answers in simplest form: (f(4 + h)=)(f(4 + h)-f(4)=)(\frac{f(4 + h)-f(4)}{h}=)(lim_{h
ightarrow0}\frac{f(4 + h)-f(4)}{h}=)so, (f(4)=)

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

ightarrow0}\frac{f(4 + h)-f(4)}{h}$
\[

$$\begin{align*} \lim_{h ightarrow0}\frac{f(4 + h)-f(4)}{h}&=\lim_{h ightarrow0}(40 + 5h)\\ &=40 \end{align*}$$

\]
Since $f^{\prime}(4)=\lim_{h
ightarrow0}\frac{f(4 + h)-f(4)}{h}$, $f^{\prime}(4)=40$.

Answer:

$f(4 + h)=80 + 40h+5h^{2}$
$f(4 + h)-f(4)=40h + 5h^{2}$
$\frac{f(4 + h)-f(4)}{h}=40 + 5h$
$\lim_{h
ightarrow0}\frac{f(4 + h)-f(4)}{h}=40$
$f^{\prime}(4)=40$