Question

the point $(-3, 6)$ is on the graph of the function $f(x) = -6sqrt3{x + 2}$ as shown.

answer the parts below to estimate the instantaneous rate of change of $f(x)$ at $x = -3$.

(a) find the average rate of change of $f(x)$ over each given interval in the table below. do not round intermediate computations, and round your answers to 4 decimal places if necessary.

interval	$-3.1, -3$	$-3.01, -3$	$-3.001, -3$	$-3, -2.999$	$-3, -2.99$	$-3, -2.9$

(b) given the graph and the table above, give the apparent instantaneous rate of change of $f(x)$ at $x = -3$.

$square$

Explanation:

The average rate of change of a function $f(x)$ over an interval $[a,b]$ is given by $\frac{f(b)-f(a)}{b-a}$. We know $f(-3)=6$.

Step1: Calculate for $[-3.1, -3]$

First find $f(-3.1)$:
$f(-3.1)=-6\sqrt[3]{-3.1+2}=-6\sqrt[3]{-1.1}=6\sqrt[3]{1.1}\approx6\times1.032280115$
Average rate of change:
$\frac{f(-3)-f(-3.1)}{-3-(-3.1)}=\frac{6 - 6\sqrt[3]{1.1}}{0.1}\approx\frac{6 - 6.19368069}{0.1}=\frac{-0.19368069}{0.1}\approx-1.9368$

Step2: Calculate for $[-3.01, -3]$

Find $f(-3.01)$:
$f(-3.01)=-6\sqrt[3]{-3.01+2}=-6\sqrt[3]{-1.01}=6\sqrt[3]{1.01}\approx6\times1.003322284$
Average rate of change:
$\frac{f(-3)-f(-3.01)}{-3-(-3.01)}=\frac{6 - 6\sqrt[3]{1.01}}{0.01}\approx\frac{6 - 6.019933704}{0.01}=\frac{-0.019933704}{0.01}\approx-1.9934$

Step3: Calculate for $[-3.001, -3]$

Find $f(-3.001)$:
$f(-3.001)=-6\sqrt[3]{-3.001+2}=-6\sqrt[3]{-1.001}=6\sqrt[3]{1.001}\approx6\times1.000333222$
Average rate of change:
$\frac{f(-3)-f(-3.001)}{-3-(-3.001)}=\frac{6 - 6\sqrt[3]{1.001}}{0.001}\approx\frac{6 - 6.001999333}{0.001}=\frac{-0.001999333}{0.001}\approx-1.9993$

Step4: Calculate for $[-3, -2.999]$

Find $f(-2.999)$:
$f(-2.999)=-6\sqrt[3]{-2.999+2}=-6\sqrt[3]{-0.999}=6\sqrt[3]{0.999}\approx6\times0.999666555$
Average rate of change:
$\frac{f(-2.999)-f(-3)}{-2.999-(-3)}=\frac{6\sqrt[3]{0.999}-6}{0.001}\approx\frac{5.99799933 - 6}{0.001}=\frac{-0.00200067}{0.001}\approx-2.0007$

Step5: Calculate for $[-3, -2.99]$

Find $f(-2.99)$:
$f(-2.99)=-6\sqrt[3]{-2.99+2}=-6\sqrt[3]{-0.99}=6\sqrt[3]{0.99}\approx6\times0.996655493$
Average rate of change:
$\frac{f(-2.99)-f(-3)}{-2.99-(-3)}=\frac{6\sqrt[3]{0.99}-6}{0.01}\approx\frac{5.979932958 - 6}{0.01}=\frac{-0.020067042}{0.01}\approx-2.0067$

Step6: Calculate for $[-3, -2.9]$

Find $f(-2.9)$:
$f(-2.9)=-6\sqrt[3]{-2.9+2}=-6\sqrt[3]{-0.9}=6\sqrt[3]{0.9}\approx6\times0.965489385$
Average rate of change:
$\frac{f(-2.9)-f(-3)}{-2.9-(-3)}=\frac{6\sqrt[3]{0.9}-6}{0.1}\approx\frac{5.79293631 - 6}{0.1}=\frac{-0.20706369}{0.1}\approx-2.0706$

Step7: Estimate instantaneous rate

As the intervals get smaller around $x=-3$, the average rates approach $-2$.

Answer:

Part (a)

Interval	Average rate of change
$[-3.01, -3]$	$-1.9934$
$[-3.001, -3]$	$-1.9993$
$[-3, -2.999]$	$-2.0007$
$[-3, -2.99]$	$-2.0067$
$[-3, -2.9]$	$-2.0706$

Part (b)

$-2$