Question

f(x)=2^{x + 2}+2

1. using the graph calculate the average rate of change in the interval -2≤x≤1

(-1,4) (1,10)
arc = \frac{10 - 4}{1-(-1)}

1. using the graph calculate the instantaneous rate of change at the point where x = 1

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, $a=-2$, $b = 1$. First, find $f(-2)$ and $f(1)$.
For $x=-2$, $f(-2)=2^{-2 + 2}+2=2^0+2=1 + 2=3$.
For $x = 1$, $f(1)=2^{1+2}+2=2^3+2=8 + 2=10$.

Step2: Calculate average rate of change

Using the formula $\frac{f(1)-f(-2)}{1-(-2)}=\frac{10 - 3}{1+2}=\frac{7}{3}$.

Step3: Recall instantaneous rate - of - change (graphical approach)

The instantaneous rate of change of a function $y = f(x)$ at a point $x = c$ is the slope of the tangent line to the graph of the function at $x = c$. To estimate it from the graph, we draw the tangent line at $x = 1$ and find its slope.
If we assume we can estimate two points on the tangent line at $x = 1$ as $(1,10)$ and $(2,18)$ (by observing the graph and making an estimate).

Step4: Calculate slope of tangent line

The slope $m=\frac{y_2-y_1}{x_2-x_1}=\frac{18 - 10}{2 - 1}=8$.

Answer:

1. $\frac{7}{3}$
2. 8