Question

f(x)=3\sqrt{(x + 3)}-2

1. using the graph calculate the average rate of change in the interval 1≤x≤6
2. using the graph calculate the instantaneous rate of change at the point where x = 6

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, $a = 1$, $b = 6$, and $f(x)=3\sqrt{x + 3}-2$.
First, find $f(1)$:
$f(1)=3\sqrt{1 + 3}-2=3\times2-2=6 - 2=4$.
Then, find $f(6)$:
$f(6)=3\sqrt{6 + 3}-2=3\times3-2=9 - 2=7$.
The average rate of change is $\frac{f(6)-f(1)}{6 - 1}=\frac{7 - 4}{5}=\frac{3}{5}=0.6$.

Step2: Recall instantaneous rate - of - change concept

The instantaneous rate of change of a function $y = f(x)$ at $x = c$ is the slope of the tangent line to the graph of the function at $x = c$.
To find the derivative of $f(x)=3\sqrt{x + 3}-2=3(x + 3)^{\frac{1}{2}}-2$.
Using the power rule and chain - rule, if $u=x + 3$, then $y = 3u^{\frac{1}{2}}-2$.
$\frac{dy}{du}=\frac{3}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=1$.
So, $f^\prime(x)=\frac{3}{2\sqrt{x + 3}}$.
When $x = 6$, $f^\prime(6)=\frac{3}{2\sqrt{6+3}}=\frac{3}{2\times3}=\frac{1}{2}=0.5$.

Answer:

1. The average rate of change over the interval $1\leq x\leq6$ is $0.6$.
2. The instantaneous rate of change at $x = 6$ is $0.5$.