Question

f(x)=x^{2}-x - 1
over which interval does f have an average rate of change of zero?
choose 1 answer:
-1≤x≤2
2≤x≤3
-5≤x≤5
-3≤x≤-2

Explanation:

Step1: Recall the average rate of change formula

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

Step2: Calculate \(f(x)\) for each option

• Option A: \(-1\leq x\leq2\)
• Calculate \(f(-1)\): \(f(-1)=(-1)^{2}-(-1)-1=1 + 1-1=1\)
• Calculate \(f(2)\): \(f(2)=2^{2}-2 - 1=4-2 - 1=1\)
• Since \(f(-1)=f(2) = 1\), the average rate of change \(\frac{f(2)-f(-1)}{2-(-1)}=\frac{1 - 1}{3}=0\)
• Option B: \(2\leq x\leq3\)
• \(f(2)=1\) (calculated above)
• \(f(3)=3^{2}-3 - 1=9-3 - 1=5\)
• \(\frac{f(3)-f(2)}{3 - 2}=\frac{5 - 1}{1}=4

eq0\)

• Option C: \(-5\leq x\leq5\)
• \(f(-5)=(-5)^{2}-(-5)-1=25 + 5-1=29\)
• \(f(5)=5^{2}-5 - 1=25-5 - 1=19\)
• \(\frac{f(5)-f(-5)}{5-(-5)}=\frac{19 - 29}{10}=-1

eq0\)

• Option D: \(-3\leq x\leq-2\)
• \(f(-3)=(-3)^{2}-(-3)-1=9 + 3-1=11\)
• \(f(-2)=(-2)^{2}-(-2)-1=4 + 2-1=5\)
• \(\frac{f(-2)-f(-3)}{-2-(-3)}=\frac{5 - 11}{1}=-6

eq0\)

Answer:

A. \(-1\leq x\leq2\)