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Given the function [tex]$f(x)=x^2-x-1$[/tex], over which interval does [tex]f[/tex] have an average rate of change of zero?

Choose 1 answer:
(A) [tex]2 \leq x \leq 3[/tex]
(B) [tex]-1 \leq x \leq 2[/tex]
(C) [tex]-5 \leq x \leq 5[/tex]
(D) [tex]-3 \leq x \leq -2[/tex]

Sagot :

GinnyAnsweridn
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]
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