Math Problem Statement

Given the function h(x) = x^2 - 10x + 21, determine the average rate of change of the function over the interval -1 ≤ x ≤ 10.

Solution

To determine the average rate of change of the function h(x)=x210x+21h(x) = x^2 - 10x + 21 over the interval 1x10-1 \leq x \leq 10, we use the formula for the average rate of change of a function over an interval [a,b][a, b]:

Average rate of change=h(b)h(a)ba\text{Average rate of change} = \frac{h(b) - h(a)}{b - a}

Here, a=1a = -1 and b=10b = 10. First, we need to evaluate h(1)h(-1) and h(10)h(10).

  1. Find h(1)h(-1): h(1)=(1)210(1)+21=1+10+21=32h(-1) = (-1)^2 - 10(-1) + 21 = 1 + 10 + 21 = 32

  2. Find h(10)h(10): h(10)=(10)210(10)+21=100100+21=21h(10) = (10)^2 - 10(10) + 21 = 100 - 100 + 21 = 21

Now, we can calculate the average rate of change:

Average rate of change=h(10)h(1)10(1)=213210+1=1111=1\text{Average rate of change} = \frac{h(10) - h(-1)}{10 - (-1)} = \frac{21 - 32}{10 + 1} = \frac{-11}{11} = -1

Thus, the average rate of change of the function over the interval 1x10-1 \leq x \leq 10 is -1.

Would you like more details or have any questions?

Here are five related questions to explore:

  1. What does the average rate of change tell us about a function's behavior?
  2. How do we interpret a negative average rate of change geometrically?
  3. What is the difference between average rate of change and instantaneous rate of change?
  4. How do you find the instantaneous rate of change using derivatives?
  5. How does the shape of a quadratic function affect its rate of change?

Tip: The average rate of change of a quadratic function over any interval depends on how the curve behaves over that range. For linear functions, the rate of change is constant over any interval.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Functions
Rate of Change

Formulas

Average rate of change = (f(b) - f(a)) / (b - a)

Theorems

Definition of the Average Rate of Change

Suitable Grade Level

Grades 10-12