Question

Hello. Im trying to help my 9th grade autisitc daughter with her test corrections. Its been over 20 yrs since I took Algebra 1 and Im a little rusty. She gets easily frustrated and so Im trying to do some of the leg work before trying to explain it to her. Thanks so much in advance . I was working with another tutor and he was helping me with part 2 but we got disconnected

Answer 1

To find the average rate of change over the given intervals, we need to remember that the average rate of change is given by:

$A_{\text{rateofchange}}=\frac{change\text{ in y}}{\text{change in x}}_{}=\frac{f(b)-f(a)}{b-a}$

We already have that the function is:

$f(x)=55(0.7)^x$

Then, if we want to find the average rate of change over the interval:

[1,3] (expressed in interval notation), we have to use the values for x = 1, and x = 3 as inputs for the function. Then, we have:

the average rate of change over

$A_{\text{rateofchange}}=\frac{f(3)-f(1)}{3-1}$

And we have that:

$f(x)=55(0.7)^x\Rightarrow f(3)=18.865$

$f(1)=38.5$

Then the rate of change in the given interval is:

$A_{\text{rateofchange}}=\frac{18.865-38.5}{3-1}=\frac{-19.635}{2}=-9.8175$

If we round the result to three decimal places, we have that the average rate of change in the interval is -9.818.

We can follow the same procedure for the interval [4, 8] as follows:

$f(x)=55(0.7)^x\Rightarrow f(8)=55(0.7)^8_{}=3.17064055$

$f(4)=55(0.7)^4\Rightarrow f(4)=13.2055$

If we apply the concept of the average rate of change, then we have:

$A_{\text{rateofchange}}=\frac{f(8)-f(4)}{8-4}=\frac{3.17064055-13.2055}{8-4}$

Then, we have:

$A_{\text{rateofchange}}=\frac{-10.03485945}{4}\Rightarrow A_{rateofchange}=-2.5087148625$

If we round the result to three decimal places, then we have that the average rate of change over the interval [4, 8] is -2.509.

Since we have that over the interval [1, 3] is -9.818, and over the interval [4, 8] the average rate of change is -2.509, we can conclude that it increases as x increases.