What is the average rate of change of f(x) = (x + 3)2 − 2

When we have a function f(x), the average rate of change in the interval (a, b) is:

In this case, we have the function:

f(x) = (x + 3)^2 - 2

(but we do not have the interval, and I couldn't find the complete question online)

So if for example, we have the interval (2, 4)

The average rate of change will be:

If instead, we want the rate of change in a differential dx around the value x, we need to differentiate the function (this is way more complex, so I will define some rules first).

Such that the rate of change, in this case, will be:

f'(x) = df/dx

For a function like:

g(x) = x^n + c

g'(x) = n*x^(n - 1)

And for:

h(x) = k( g(x))

h'(x) = k'(g(x))*g'(x)

So here we can write our function as:

f(x) = k(g(x)) = (x + 3)^2 - 2

where:

g(x) = x + 3

k(x) = x^2 - 2

Then:

f'(x) = 2*(x + 3)*1 = 2*x + 6

That is the rate of change as a function of x (but is not an "average" rate of change)

b. The student's scores on the posttest would have a smaller standard deviation.

Step-by-step explanation:

The first test is taken before the material is covered in class so we expect the standard deviation to be high because not everyone's scores would be lying close to the mean. Equal number of students mastered most, some or almost none of the material from reading the textbook based on the pretest result. this means the data is varying, so the standard deviation is large.

Whereas, after the teacher has taught the material and given the homework, they must have understood most of the material. The test they take after teaching as a post test. The results of the post test would have a smaller standard deviation as most of the students would have scored good. Hence, the student's scores on the posttest would have a smaller standard deviation.