f(x)=3\\sqrt{(x + 3)}-2\n1. using the graph calculate the average rate of change in the interval…

f(x)=3\\sqrt{(x + 3)}-2\n1. using the graph calculate the average rate of change in the interval 1≤x≤6\n(1,4) 3\\sqrt{(1 + 3)}-2=4 (6,7) 3\\sqrt{(6 + 3)}-2=7 arcc=\\frac{7 - 4}{6 - 1}=0.6\n2. using the graph calculate the instantaneous rate of change at the point where x = 6

f(x)=3\\sqrt{(x + 3)}-2\n1. using the graph calculate the average rate of change in the interval 1≤x≤6\n(1,4) 3\\sqrt{(1 + 3)}-2=4 (6,7) 3\\sqrt{(6 + 3)}-2=7 arcc=\\frac{7 - 4}{6 - 1}=0.6\n2. using the graph calculate the instantaneous rate of change at the point where x = 6

Answer

Explanation:

Step1: Recall the formula for instantaneous rate of change

The instantaneous rate of change of a function $y = f(x)$ at a point $x = a$ is given by the slope of the tangent line to the graph of the function at $x=a$. Without the actual graph - we can also use the derivative. First, find the derivative of $f(x)=3\sqrt{x + 3}-2=3(x + 3)^{\frac{1}{2}}-2$.

Step2: Apply the power - rule for differentiation

The power - rule states that if $y = u^n$, then $y^\prime=nu^{n - 1}u^\prime$. Let $u=x + 3$, $n=\frac{1}{2}$. Then $f^\prime(x)=3\times\frac{1}{2}(x + 3)^{\frac{1}{2}-1}\times1=\frac{3}{2\sqrt{x + 3}}$.

Step3: Evaluate the derivative at $x = 6$

Substitute $x = 6$ into $f^\prime(x)$. We have $f^\prime(6)=\frac{3}{2\sqrt{6+3}}=\frac{3}{2\times3}=\frac{1}{2}=0.5$.

Answer:

$0.5$