Math Problem Statement

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The following is a partial table of values for f(x). Find the following: 1. What is the value of f(10)? 2. Calculate the average rate of change from x = 5 to x = 10. Use your answer to find f(7).

Solution

Let's break down the two parts of the question based on the table provided in your image.

Part a. What is the value of f(10)f(10)?

From the table, you can directly read off the value of f(10)f(10).

  • When x=10x = 10, f(10)=1.1f(10) = 1.1.

Part b. Calculate the average rate of change from x=5x = 5 to x=10x = 10. Use your answer to find f(7)f(7).

The average rate of change between two points is calculated using the formula:

Average rate of change=f(x2)f(x1)x2x1\text{Average rate of change} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}

For x1=5x_1 = 5 and x2=10x_2 = 10:

Average rate of change=f(10)f(5)105\text{Average rate of change} = \frac{f(10) - f(5)}{10 - 5}

Substitute the values from the table:

f(10)=1.1andf(5)=4.3f(10) = 1.1 \quad \text{and} \quad f(5) = 4.3

Now calculate:

Average rate of change=1.14.3105=3.25=0.64\text{Average rate of change} = \frac{1.1 - 4.3}{10 - 5} = \frac{-3.2}{5} = -0.64

So, the average rate of change is 0.64-0.64.

Use this to estimate f(7)f(7).

Since 7 is between 5 and 10, we can assume a linear change between those points. To find f(7)f(7), we'll use the slope m=0.64m = -0.64 and apply the point-slope form of the line between x=5x = 5 and x=10x = 10. Let's take the point (5,4.3)(5, 4.3):

f(7)=f(5)+m(x5)f(7) = f(5) + m(x - 5)

Substitute x=7x = 7, f(5)=4.3f(5) = 4.3, and m=0.64m = -0.64:

f(7)=4.3+(0.64)(75)=4.3+(0.64)(2)=4.31.28=3.02f(7) = 4.3 + (-0.64)(7 - 5) = 4.3 + (-0.64)(2) = 4.3 - 1.28 = 3.02

So, f(7)3.02f(7) \approx 3.02.

Summary:

  • f(10)=1.1f(10) = 1.1
  • Average rate of change between x=5x = 5 and x=10x = 10 is 0.64-0.64
  • f(7)3.02f(7) \approx 3.02

Would you like any additional details on the process?

Here are 5 related questions to further your understanding:

  1. How would the calculation change if the points between 5 and 10 were not linear?
  2. How can you use the concept of the average rate of change in real-world problems?
  3. Can you calculate the rate of change between x=0x = 0 and x=5x = 5?
  4. How does the slope of a function inform us about its behavior over an interval?
  5. What would f(12)f(12) be assuming the same average rate of change applies beyond x=10x = 10?

Tip: When dealing with linear approximations, the assumption is that the rate of change remains constant within the interval, but this might not always be true for non-linear functions!

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Math Problem Analysis

Mathematical Concepts

Functions
Average Rate of Change
Linear Approximation

Formulas

Average rate of change formula: (f(x2) - f(x1)) / (x2 - x1)
Linear approximation using point-slope form

Theorems

Point-Slope Form Theorem
Linear Interpolation

Suitable Grade Level

Grades 9-10