To determine over which interval the function [tex]\( f(x) = x^2 - x - 1 \)[/tex] has an average rate of change of zero, we need to follow these steps:
1. We start with the definition of the average rate of change of a function [tex]\( f \)[/tex] over an interval [tex]\([a, b]\)[/tex]. It is given by the formula:
[tex]\[
\text{average rate of change} = \frac{f(b) - f(a)}{b - a}
\][/tex]
2. Let's examine each of the given intervals to find the average rate of change in each case.
- Interval (A) [tex]\( 2 \leq x \leq 3 \)[/tex]:
[tex]\[
\text{average rate of change} = \frac{f(3) - f(2)}{3 - 2}
\][/tex]
[tex]\[
= \frac{((3)^2 - 3 - 1) - ((2)^2 - 2 - 1)}{3 - 2} = \frac{(9 - 3 - 1) - (4 - 2 - 1)}{1} = \frac{5 - 1}{1} = 4
\][/tex]
- Interval (B) [tex]\( -1 \leq x \leq 2 \)[/tex]:
[tex]\[
\text{average rate of change} = \frac{f(2) - f(-1)}{2 - (-1)}
\][/tex]
[tex]\[
= \frac{((2)^2 - 2 - 1) - ((-1)^2 - (-1) - 1)}{2 - (-1)} = \frac{(4 - 2 - 1) - (1 + 1 - 1)}{3} = \frac{1 - 1}{3} = 0
\][/tex]
- Interval (C) [tex]\( -5 \leq x \leq 5 \)[/tex]:
[tex]\[
\text{average rate of change} = \frac{f(5) - f(-5)}{5 - (-5)}
\][/tex]
[tex]\[
= \frac{((5)^2 - 5 - 1) - ((-5)^2 - (-5) - 1)}{5 - (-5)} = \frac{(25 - 5 - 1) - (25 + 5 - 1)}{10} = \frac{19 - 29}{10} = -1
\][/tex]
- Interval (D) [tex]\( -3 \leq x \leq -2 \)[/tex]:
[tex]\[
\text{average rate of change} = \frac{f(-2) - f(-3)}{-2 - (-3)}
\][/tex]
[tex]\[
= \frac{((-2)^2 - (-2) - 1) - ((-3)^2 - (-3) - 1)}{-2 - (-3)} = \frac{(4 + 2 - 1) - (9 + 3 - 1)}{1} = \frac{5 - 11}{1} = -6
\][/tex]
From the above calculations, we can see that the average rate of change of the function is zero only over the interval [tex]\( -1 \leq x \leq 2 \)[/tex].
Therefore, the correct answer is:
[tex]\[
\boxed{-1 \leq x \leq 2}
\][/tex]
