Find Average rate of change of the function f(x)=x^4-5x on the closed interval [0,3]?

The **average rate of change** of a function \( f(x) \) over a closed interval \([a, b]\) is given by the formula:

\[
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
\]

For the function \( f(x) = x^4 - 5x \) on the interval \([0, 3]\), we need to calculate:

1. \( f(0) \)
2. \( f(3) \)
3. Plug the values into the formula.

### Step 1: Calculate \( f(0) \)
\[
f(0) = 0^4 - 5(0) = 0
\]

### Step 2: Calculate \( f(3) \)
\[
f(3) = 3^4 - 5(3) = 81 - 15 = 66
\]

### Step 3: Use the average rate of change formula
\[
\text{Average Rate of Change} = \frac{f(3) - f(0)}{3 - 0} = \frac{66 - 0}{3 - 0} = \frac{66}{3} = 22
\]

Thus, the average rate of change of the function \( f(x) = x^4 - 5x \) on the interval \([0, 3]\) is **22**.

Do you want more details or have any questions?

Here are 5 related questions to deepen your understanding:
1. How do you find the instantaneous rate of change of \( f(x) = x^4 - 5x \) at \( x = 3 \)?
2. What is the average rate of change of \( f(x) = x^4 - 5x \) on the interval \([1, 3]\)?
3. How does the average rate of change differ from the derivative at a point?
4. How would the average rate of change change if the interval was \([-1, 3]\)?
5. What is the significance of a positive average rate of change in terms of the behavior of the function?

**Tip**: The average rate of change over an interval is the slope of the secant line connecting the two points on the function graph over that interval.

Find the average rate of change of the function f(x)=x^4-5x on the closed interval [0,3]?

Discover how to calculate the average rate of change for the function f(x)=x^4-5x on the interval [0,3]. This solution covers key calculus concepts such as average rate of change and polynomial functions. Suitable for students in Grades 10-12.

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