Math Problem Statement

Let f be defined as follows: y = f(x) = x^2 - 2x. (a) Find the average rate of change of y with respect to x in the following intervals: from x = 2 to x = 3, from x = 2 to x = 2.5, and from x = 2 to x = 2.1. (b) Find the instantaneous rate of change of y at x = 2.

Solution

Let's work through the problem you've uploaded.

Given that:

f(x)=x22xf(x) = x^2 - 2x

Part (a) - Finding the average rate of change

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

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

1. From x=2x = 2 to x=3x = 3:

We need to compute f(3)f(3) and f(2)f(2):

f(3)=322(3)=96=3f(3) = 3^2 - 2(3) = 9 - 6 = 3 f(2)=222(2)=44=0f(2) = 2^2 - 2(2) = 4 - 4 = 0

The average rate of change is:

f(3)f(2)32=3032=3\frac{f(3) - f(2)}{3 - 2} = \frac{3 - 0}{3 - 2} = 3

2. From x=2x = 2 to x=2.5x = 2.5:

We compute f(2.5)f(2.5):

f(2.5)=(2.5)22(2.5)=6.255=1.25f(2.5) = (2.5)^2 - 2(2.5) = 6.25 - 5 = 1.25

The average rate of change is:

f(2.5)f(2)2.52=1.2502.52=1.250.5=2.5\frac{f(2.5) - f(2)}{2.5 - 2} = \frac{1.25 - 0}{2.5 - 2} = \frac{1.25}{0.5} = 2.5

3. From x=2x = 2 to x=2.1x = 2.1:

We compute f(2.1)f(2.1):

f(2.1)=(2.1)22(2.1)=4.414.2=0.21f(2.1) = (2.1)^2 - 2(2.1) = 4.41 - 4.2 = 0.21

The average rate of change is:

f(2.1)f(2)2.12=0.2102.12=0.210.1=2.1\frac{f(2.1) - f(2)}{2.1 - 2} = \frac{0.21 - 0}{2.1 - 2} = \frac{0.21}{0.1} = 2.1

Part (b) - Instantaneous rate of change at x=2x = 2

The instantaneous rate of change is given by the derivative f(x)f'(x).

First, we compute the derivative of f(x)=x22xf(x) = x^2 - 2x:

f(x)=2x2f'(x) = 2x - 2

At x=2x = 2, we get:

f(2)=2(2)2=42=2f'(2) = 2(2) - 2 = 4 - 2 = 2

Final Answers:

  • (a) Average rates of change:
    1. From x=2x = 2 to x=3x = 3: 33
    2. From x=2x = 2 to x=2.5x = 2.5: 2.52.5
    3. From x=2x = 2 to x=2.1x = 2.1: 2.12.1
  • (b) Instantaneous rate of change at x=2x = 2: 22

Would you like further clarification or details on any part?

Here are 5 related questions to expand on:

  1. How does the average rate of change differ from the instantaneous rate of change?
  2. Can the derivative give us information about concavity of the function?
  3. How would the average rate of change change for other intervals around x=2x = 2?
  4. What happens to the rate of change as xx approaches 2 from either side?
  5. How can we visualize the derivative graphically?

Tip: The average rate of change between two points on a curve is the slope of the secant line connecting them. The instantaneous rate of change at a point is the slope of the tangent line at that point.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Average Rate of Change
Instantaneous Rate of Change
Derivatives

Formulas

f(x) = x^2 - 2x
Average rate of change formula: (f(b) - f(a)) / (b - a)
Derivative of f(x): f'(x) = 2x - 2

Theorems

Derivative as Instantaneous Rate of Change

Suitable Grade Level

Grades 10-12