By definition, the average rate of change is given by:
[tex] AVR = \frac{f(x2)-f(x1)}{x2-x1} [/tex]
We evaluate each of the functions in the given interval.
We have then:
For f (x) = x ^ 2 + 3x:
Evaluating for x = -2:
[tex] f (-2) = (-2) ^ 2 + 3 (-2)
f (-2) = 4 - 6
f (-2) = - 2
[/tex]
Evaluating for x = 3:
[tex] f (3) = (3) ^ 2 + 3 (3)
f (3) = 9 + 9
f (3) = 18
[/tex]
Then, the AVR is:
[tex] AVR = \frac{18-(-2)}{3-(-2)} [/tex]
[tex] AVR = \frac{18+2}{3+2} [/tex]
[tex] AVR = \frac{20}{5} [/tex]
[tex] AVR = 4 [/tex]
For f (x) = 3x - 8:
Evaluating for x =4:
[tex] f (4) = 3 (4) - 8
f (4) = 12 - 8
f (4) = 4
[/tex]
Evaluating for x = 5:
[tex] f (5) = 3 (5) - 8
f (5) = 15 - 8
f (5) = 7 [/tex]
Then, the AVR is:
[tex] AVR = \frac{7-4}{5-4} [/tex]
[tex] AVR = \frac{3}{1} [/tex]
[tex] AVR = 3 [/tex]
For f (x) = x ^ 2 - 2x:
Evaluating for x = -3:
[tex] f (-3) = (-3) ^ 2 - 2 (-3)
f (-3) = 9 + 6
f (-3) = 15
[/tex]
Evaluating for x = 4:
[tex] f (4) = (4) ^ 2 - 2 (4)
f (4) = 16 - 8
f (4) = 8
[/tex]
Then, the AVR is:
[tex] AVR = \frac{8-15}{4-(-3)} [/tex]
[tex] AVR = \frac{-7}{4+3} [/tex]
[tex] AVR = \frac{-7}{7} [/tex]
[tex] AVR = -1 [/tex]
For f (x) = x ^ 2 - 5:
Evaluating for x = -1:
[tex] f (-1) = (-1) ^ 2 - 5
f (-1) = 1 - 5
f (-1) = - 4
[/tex]
Evaluating for x = 1:
[tex] f (1) = (1) ^ 2 - 5
f (1) = 1 - 5
f (1) = - 4
[/tex]
Then, the AVR is:
[tex] AVR = \frac{-4-(-4)}{1-(-1)} [/tex]
[tex] AVR = \frac{-4+4}{1+1} [/tex]
[tex] AVR = \frac{0}{2} [/tex]
[tex] AVR = 0 [/tex]
Answer:
from the greatest to the least value based on the average rate of change in the specified interval:
f(x) = x^2 + 3x interval: [-2, 3]
f(x) = 3x - 8 interval: [4, 5]
f(x) = x^2 - 5 interval: [-1, 1]
f(x) = x^2 - 2x interval: [-3, 4]
