Answer::
Average rate of change(A(x)) of f(x) over the interval [a, b] is given by:
[tex]A(x) = \frac{f(b)-f(a)}{b-a}[/tex]
As per the statement:
Given :
f(x) = x+4 over the interval [4, 5]
At x = 4
f(4) = 8
at x = 5
f(5) = 9
Using the above formula: we have;
[tex]A_1(x) = \frac{f(5)-f(4)}{5-4} = \frac{9-8}{1} = \frac{1}{1} = 1[/tex]
Similarly for :
f(x) = 3x-7 over the interval [-3, 7]
at x = -3
f(-3) = -16
At x = 7:
f(7) = 14
then;
[tex]A_2(x) = \frac{f(7)-f(-3)}{7-(-3)} = \frac{14+16}{10} = \frac{30}{10} = 3[/tex]
For:
f(x) = 9x + 1 over the interval [-5, -2]
At x = -5
f(-5) = -44
At x = -2
f(-2) = -17
then;
[tex]A_3(x) = \frac{f(-2)-f(-5)}{-2-(-5)} = \frac{-17+44}{3} = \frac{27}{3} =9[/tex]
For:
f(x) = 2x + 9 over the interval [1, 2]
At x = 1
f(1) = 11
At x =2
f(2)= 13
then;
[tex]A_4(x) = \frac{f(2)-f(1)}{2-1} = \frac{13-11}{1} = \frac{2}{1} =2[/tex]
We have to Arrange these functions from the least to the greatest value based on the average rate of change in the specified interval.
[tex]A_1 < A_4< A_2<A_3[/tex]
⇒[tex]1< 2< 3< 9[/tex]
⇒x+4 < 2x + 9 < 3x-7 < 9x + 1
Therefore, the least to the greatest value based on the average rate of change in the specified interval is:
f(x) = x+4, f(x) = 2x+9, f(x) = 3x-7, f(x) = 9x+1,
