a) The average rate of change over the interval -1 ≤ x <1 =b) The average rate of change over the interval 1 ≤ x < 3 =c) The average rate of change over the interval 3 ≤ x ≤ 5=

a) The average rate of change over the interval -1 ≤ x <1 =b) The average rate of change over the interval 1 ≤ x < 3 =c) The average rate of change over the int class=

Answer :

Hello there. To solve this question, we'll have to remember how to determine the average rate of change of a function over an interval.

Given a function f(x), over an interval [a, b], the average rate of change, noted by avg(f) can be calculated by the formula:

[tex]\mathrm{avg}(f)=\dfrac{f(b)-f(a)}{b-a}[/tex]

We already know the length of the intervals are equal to 2, which means that we're simply taking the arithmetic mean of the interval extrema.

Let's start calculating the average rate of change of the function f(x) = 4 - x² over the intervals:

-1 ≤ x ≤ 1

In this case, a = -1 and b = 1. Plugging the values, we get:

[tex]\mathrm{avg}(f)=\dfrac{4-1^2-(4-(-1)^2)}{2}[/tex]

Calculate the squares and add the values

[tex]\mathrm{avg}(f)=\dfrac{4-1-(4-1)}{2}=\frac{4-1-4+1}{2}=\frac{0}{2}=0[/tex]

This means that the average rate of change of this function is equal to zero over this interval.

1 ≤ x ≤ 3

Plugging the values, we get:

[tex]\mathrm{avg}(f)=\dfrac{4-3^2-(4-1^2)}{2}[/tex]

Calculate the squares and add the values

[tex]\mathrm{avg}(f)=\dfrac{4-9-(4-1)}{2}=\frac{4-9-4+1}{2}=\frac{-8}{2}=-4[/tex]

This means that the function may be decreasing over this interval. Remember the average rate of change is also the slope of the secant line of the function and when it is negative, it means the function is also decreasing).

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