Questions: Compare the rate of change for the function f(x)=x^3-2x^2+x+1 over the intervals [0,2] and [2,4]

Solution Steps

To compare the rate of change of the function $f(x) = x^3 - 2x^2 + x + 1$ over the intervals $[0,2]$ and $[2,4]$, we need to calculate the average rate of change for each interval. The average rate of change of a function $f(x)$ over an interval $[a, b]$ is given by the formula $\frac{f(b) - f(a)}{b - a}$. We will apply this formula to both intervals and compare the results.

To find the average rate of change of the function $f(x) = x^3 - 2x^2 + x + 1$ over the interval $[0, 2]$, we use the formula:

$$\text{Average Rate of Change} = \frac{f(2) - f(0)}{2 - 0}$$

Calculating $f(2)$ and $f(0)$:

$$f(2) = 2^3 - 2(2^2) + 2 + 1 = 8 - 8 + 2 + 1 = 3$$ $$f(0) = 0^3 - 2(0^2) + 0 + 1 = 1$$

Thus, the average rate of change over $[0, 2]$ is:

$$\frac{3 - 1}{2 - 0} = \frac{2}{2} = 1.0$$

Next, we calculate the average rate of change over the interval $[2, 4]$ using the same formula:

$$\text{Average Rate of Change} = \frac{f(4) - f(2)}{4 - 2}$$

Calculating $f(4)$:

$$f(4) = 4^3 - 2(4^2) + 4 + 1 = 64 - 32 + 4 + 1 = 37$$

Thus, the average rate of change over $[2, 4]$ is:

$$\frac{37 - 3}{4 - 2} = \frac{34}{2} = 17.0$$

We have found the following average rates of change:

• Over $[0, 2]$: $1.0$
• Over $[2, 4]$: $17.0$

Since $17.0 > 1.0$, the average rate of change is much greater over the interval $[2, 4]$.

Final Answer