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Over what intervals is the average rate of change of f(x) = 5x greater than the average rate of change of g(x) = 25x? Select all that apply. A. 0 ≤ x ≤ 3 B. 0 ≤ x ≤ 2 C. 3 ≤ x ≤ 6 D. 1 ≤ x ≤ 2

Sagot :

MrRoyal

The average rates of change of a function f(x) and g(x) are their slopes

The average rate of change of the function f(x) is always less than the average rate of change of the function g(x)

How to determine the average rate of change?

The average rate of change of a function f(x) over the interval [a,b] is calculated as:

[tex]m = \frac{f(b) - f(a)}{b - a}[/tex]

The average rates of change of the functions over the intervals are:

A. 0 ≤ x ≤ 3

[tex]m_1 = \frac{f(3) - f(0)}{3 - 0} = \frac{5 * 3 - 5 *0}{3 - 0} = 5[/tex] ---- f(x)

[tex]m_2 = \frac{g(3) - g(0)}{3 - 0} = \frac{25 * 3 - 25 *0}{3 - 0} = 25[/tex] --- g(x)

B. 0 ≤ x ≤ 2

[tex]m_1 = \frac{f(2) - f(0)}{2 - 0} = \frac{5 * 2 - 5 *0}{2 - 0} = 5[/tex] ---- f(x)

[tex]m_2 = \frac{g(2) - g(0)}{2 - 0} = \frac{25 * 2 - 25 *0}{2 - 0} = 25[/tex] --- g(x)

C. 3 ≤ x ≤ 6

[tex]m_1 = \frac{f(6) - f(3)}{6 - 3} = \frac{5 * 6 - 5 *3}{6 - 3} = 5[/tex] ---- f(x)

[tex]m_2 = \frac{g(6) - g(3)}{6 - 3} = \frac{25 * 6 - 25 *3}{6 - 3} = 25[/tex] --- g(x)

D. 1 ≤ x ≤ 2

[tex]m_1 = \frac{f(2) - f(1)}{2 - 1} = \frac{5 * 2 - 5 *1}{2 - 1} = 5[/tex] ---- f(x)

[tex]m_2 = \frac{g(2) - g(1)}{2 - 1} = \frac{25 * 2 - 25 *1}{2 - 1} = 25[/tex] --- g(x)

From the above computation, we can see that:

The average rate of change of the function f(x) = 5x is always less than the average rate of change of the function g(x) = 25x

Read more about average rates of change at:

https://brainly.com/question/8728504

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