Question

compare the rate of change for the function ( f(x) = x^3 - 2x^2 + x + 1 ) over the intervals (0, 2) and (2, 4).
the average rate of change over the interval (0, 2) is 1. the average rate of change over the interval (2, 4) is 17. therefore, the average rate of change is much greater over the interval (2,4).

Explanation:

Step1: Recall the average rate of change formula

The average rate of change of a function \(f(x)\) over the interval \([a, b]\) is given by \(\frac{f(b)-f(a)}{b - a}\).

Step2: Calculate for \([0, 2]\)

First, find \(f(0)\) and \(f(2)\) for \(f(x)=x^{3}-2x^{2}+x + 1\).

• \(f(0)=(0)^{3}-2(0)^{2}+0 + 1=1\).
• \(f(2)=(2)^{3}-2(2)^{2}+2 + 1=8-8 + 2+1 = 3\).

Then, use the formula: \(\frac{f(2)-f(0)}{2 - 0}=\frac{3 - 1}{2}=\frac{2}{2}=1\).

Step3: Calculate for \([2, 4]\)

Find \(f(4)\): \(f(4)=(4)^{3}-2(4)^{2}+4 + 1=64-32 + 4+1 = 37\).
Use the formula: \(\frac{f(4)-f(2)}{4 - 2}=\frac{37 - 3}{2}=\frac{34}{2}=17\).

Step4: Compare the two rates

Since \(17>1\), the average rate of change is much greater over the interval \([2, 4]\).

Answer:

The average rate of change over \([0, 2]\) is \(1\), over \([2, 4]\) is \(17\), so it's greater over \([2, 4]\).