Question

question 4
use the limit definition of the derivative to find the instantaneous rate of change of $f(x)=7x^{2}+7x+4$ at $x=4$
0/1 pt つ 5 ⇄ 19 ⓘ

Explanation:

Step1: Recall limit definition of derivative

The instantaneous rate of change at $x=a$ is given by:
$$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$
Here, $a=4$, $f(x)=7x^2+7x+4$.

Step2: Compute $f(4+h)$ and $f(4)$

First, calculate $f(4+h)$:

$$\begin{align*} f(4+h)&=7(4+h)^2 +7(4+h)+4\\ &=7(16+8h+h^2)+28+7h+4\\ &=112+56h+7h^2+28+7h+4\\ &=7h^2+63h+144 \end{align*}$$

Then calculate $f(4)$:

$$\begin{align*} f(4)&=7(4)^2+7(4)+4\\ &=7(16)+28+4\\ &=112+28+4=144 \end{align*}$$

Step3: Substitute into the limit formula

$$\begin{align*} f'(4)&=\lim_{h \to 0} \frac{(7h^2+63h+144)-144}{h}\\ &=\lim_{h \to 0} \frac{7h^2+63h}{h} \end{align*}$$

Step4: Simplify the expression

Factor out $h$ in the numerator:

$$\begin{align*} f'(4)&=\lim_{h \to 0} \frac{h(7h+63)}{h}\\ &=\lim_{h \to 0} (7h+63) \end{align*}$$

Step5: Evaluate the limit

Substitute $h=0$ into the simplified expression:
$$\lim_{h \to 0} (7h+63)=7(0)+63=63$$

Answer:

63