f(x)=-4(x - 2)^2+4\n1. using the graph calculate the average rate of change in the interval 0≤x≤1\n2. using…

Explanation:

Step1: Recall average rate - of - change formula

The average rate of change of a function $y = f(x)$ over the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$. Here, $a = 0$, $b = 1$, and $f(x)=-4(x - 2)^2+4$. First, find $f(0)$: [ \begin{align*} f(0)&=-4(0 - 2)^2+4\ &=-4\times4 + 4\ &=-16 + 4\ &=-12 \end{align*} ] Then, find $f(1)$: [ \begin{align*} f(1)&=-4(1 - 2)^2+4\ &=-4\times(-1)^2+4\ &=-4 + 4\ &=0 \end{align*} ]

Step2: Calculate the average rate of change

The average rate of change over the interval $[0,1]$ is $\frac{f(1)-f(0)}{1 - 0}=\frac{0-(-12)}{1}=12$.

Step3: Recall instantaneous rate - of - change concept

The instantaneous rate of change of a function $y = f(x)$ at a point $x = c$ is the slope of the tangent line at $x = c$. First, expand the function $f(x)=-4(x - 2)^2+4=-4(x^{2}-4x + 4)+4=-4x^{2}+16x-16 + 4=-4x^{2}+16x-12$. The derivative of $f(x)$ using the power rule $(x^n)^\prime=nx^{n - 1}$ is $f^\prime(x)=-8x + 16$.

Step4: Calculate the instantaneous rate of change at $x = 3$

Substitute $x = 3$ into $f^\prime(x)$: $f^\prime(3)=-8\times3+16=-24 + 16=-8$.

Answer:

1. The average rate of change in the interval $0\leq x\leq1$ is $12$.
2. The instantaneous rate of change at $x = 3$ is $-8$.