f(x) = x² - x - 1\nover which interval does f have an average rate of change of zero?\nchoose 1 answer:\na…

Explanation:

Step1: Recall average rate of change formula

The average rate of change of a function ( f(x) ) over the interval ([a, b]) is given by (\frac{f(b) - f(a)}{b - a}). We need this to be zero, so (\frac{f(b) - f(a)}{b - a}=0), which implies ( f(b)=f(a) ).

Step2: Analyze the function ( f(x)=x^2 - x - 1 )

This is a quadratic function with ( a = 1 ), ( b=-1 ), ( c = -1 ). The axis of symmetry is ( x=-\frac{b}{2a}=-\frac{-1}{2\times1}=\frac{1}{2} ). A quadratic function is symmetric about its axis of symmetry, so ( f\left(\frac{1}{2}+h\right)=f\left(\frac{1}{2}-h\right) ) for any ( h ). We can also check each interval:

• Option A: ([-1, 2]) Calculate ( f(-1)=(-1)^2-(-1)-1 = 1 + 1 - 1=1 ) Calculate ( f(2)=2^2 - 2 - 1=4 - 2 - 1 = 1 ) Since ( f(-1)=f(2) = 1 ), then (\frac{f(2)-f(-1)}{2 - (-1)}=\frac{1 - 1}{3}=0)

• Option B: ([-5, 5]) ( f(-5)=(-5)^2-(-5)-1 = 25 + 5 - 1 = 29 ) ( f(5)=5^2 - 5 - 1=25 - 5 - 1 = 19 ) ( f(-5)\neq f(5) ), so average rate of change is not zero.

• Option C: ([-3, -2]) ( f(-3)=(-3)^2-(-3)-1 = 9 + 3 - 1 = 11 ) ( f(-2)=(-2)^2-(-2)-1 = 4 + 2 - 1 = 5 ) ( f(-3)\neq f(-2) ), so average rate of change is not zero.

• Option D: ([2, 3]) ( f(2)=1 ) (from earlier) ( f(3)=3^2 - 3 - 1=9 - 3 - 1 = 5 ) ( f(2)\neq f(3) ), so average rate of change is not zero.

Answer:

A. (-1 \leq x \leq 2)