Question

on which interval does the function $f(x) = 2x^2 - x$ have an average rate of change equal to 3?
$x = -7$ to $x = -3$
$x = -5$ to $x = -2$
$x = 0$ to $x = 2$
$x = 4$ to $x = 5$

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 to calculate this for each interval and check which one equals 3.

Step2: Analyze interval \( x = -7 \) to \( x = -3 \)

First, find \( f(-7) \) and \( f(-3) \):
\( f(-7)=2(-7)^2 - (-7)=2(49)+7 = 98 + 7=105 \)
\( f(-3)=2(-3)^2 - (-3)=2(9)+3 = 18 + 3 = 21 \)
Average rate of change: \(\frac{f(-3)-f(-7)}{-3 - (-7)}=\frac{21 - 105}{4}=\frac{-84}{4}=-21
eq3\)

Step3: Analyze interval \( x = -5 \) to \( x = -2 \)

Find \( f(-5) \) and \( f(-2) \):
\( f(-5)=2(-5)^2 - (-5)=2(25)+5 = 50 + 5 = 55 \)
\( f(-2)=2(-2)^2 - (-2)=2(4)+2 = 8 + 2 = 10 \)
Average rate of change: \(\frac{f(-2)-f(-5)}{-2 - (-5)}=\frac{10 - 55}{3}=\frac{-45}{3}=-15
eq3\)

Step4: Analyze interval \( x = 0 \) to \( x = 2 \)

Find \( f(0) \) and \( f(2) \):
\( f(0)=2(0)^2 - 0 = 0 \)
\( f(2)=2(2)^2 - 2 = 8 - 2 = 6 \)
Average rate of change: \(\frac{f(2)-f(0)}{2 - 0}=\frac{6 - 0}{2}=\frac{6}{2}=3\)

(We can check the last interval for completeness, but since we found the one that works, we can conclude)

Answer:

\( x = 0 \) to \( x = 2 \)