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For which function is the average rate of change over the interval 1 < x < 5 greater than the average rate of change over the same interval for the function g(x) = 1.8x^2?
A.=f(x) = x^2
B.=g(x) = 1.2x^2
C.=h(x) = 1.5x^2
D.=k(x) = 2x^2

Answer :

Answer:

D.=k(x) = 2x^2

Step-by-step explanation:

Average rate of change:

Given a function y, the average rate of change S of [tex]y=f(x)[/tex] in an interval  [tex](x_{s}, x_{f})[/tex] will be given by the following equation:

[tex]S = \frac{f(x_{f}) - f(x_{s})}{x_{f} - x_{s}}[/tex]

In this problem, we have that:

Interval (1,5), so [tex]x_{f} = 5, x_{s} = 1[/tex]

Then

[tex]S = \frac{f(5) - f(1)}{4}[/tex]

All options are compared to g(x).

[tex]g(x) = 1.8x^{2}[/tex]

g(5) = 1.8*(5)^2 = 45

g(1) = 1.8*(1)^2 = 1.8

[tex]S = \frac{45 - 1.8}{4} = 10.8[/tex]

Which one has a greater average rate of change?

Whichever has an average rate of change greater than 10.8.

A.=f(x) = x^2

f(5) = 5^2 = 25

f(1) = 1^2 = 1

[tex]S = \frac{25 - 1}{4} = 6.25[/tex]

Smaller than 10.8, so this is not the answer.

B.=g(x) = 1.2x^2

g(5) = 1.2*(5)^2 = 30

g(1) = 1.2*(1)^2 = 1.2

[tex]S = \frac{30 - 1.2}{4} = 7.2[/tex]

Smaller than 10.8, so this is not the answer.

C.=h(x) = 1.5x^2

h(5) = 1.5*(5)^2 = 37.5

h(1) = 1.5*(1)^2 = 1.5

[tex]S = \frac{37.5 - 1.5}{4} = 9[/tex]

Smaller than 10.8, so this is not the answer.

D.=k(x) = 2x^2

k(5) = 2*(5)^2 = 50

k(1) = 2*(1)^2 = 2

[tex]S = \frac{50 - 2}{4} = 12[/tex]

Greater than 10.8, this is the answer.

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