Problem 109 Determine the domain and estimat... [FREE SOLUTION]

Chapter 7: Problem 109

Determine the domain and estimate the range of each function. $$ r(x)=\frac{1}{x^{2}}+\frac{1}{(x-1)^{2}} $$

Short Answer

Expert verified

Domain: $ x eq 0 \text{ and } x eq 1 $. Range: $ [8, \infty) $.

Step by step solution

- Determine Points of Discontinuity

Identify the values of x that make the function undefined. For the function $ r(x) = \frac{1}{x^2} + \frac{1}{(x-1)^2} $, the denominators $ x^2 $ and $ (x-1)^2 $ should not be equal to zero. Thus, solve $ x^2 = 0 $ and $ (x-1)^2 = 0 $.

- Solve for Undefined Points

Solve the equations from Step 1: $ x^2 = 0 $ gives $ x = 0 $ and $ (x-1)^2 = 0 $ gives $ x = 1 $. Hence, the function r(x) is undefined at x = 0 and x = 1.

- State the Domain

The domain of the function r(x) excludes the points where the function is undefined. Therefore, the domain is all real numbers except x = 0 and x = 1, written as $ \text{Domain: } x eq 0 \text{ and } x eq 1 $. In interval notation, this is $ (-\infty, 0) \, \cup \, (0, 1) \, \cup \, (1, \, \infty) $.

- Analyze the Behavior at Discontinuities

Examine the behavior of the function near the points of discontinuity at x = 0 and x = 1. As x approaches these points, each term ($ 1/x^2 $ and $ 1/(x-1)^2 $) becomes very large, tending towards infinity.

- Determine the Minimum Value

Since both terms $ 1/x^2 $ and $ 1/(x-1)^2 $ are always positive, their sum $ r(x) $ is always positive. The minimum value of $ r(x) $ occurs where the sum of reciprocals is smallest. This happens midway between 0 and 1: at x = 0.5. Calculate $ r(0.5) = \frac{1}{(0.5)^2} + \frac{1}{(0.5-1)^2} = 4 + 4 = 8 $.

- State the Range

Based on the analysis, as the function approaches the points of discontinuity, the values tend to infinity. The minimum positive value reached is 8 when x = 0.5. Therefore, the range of $ r(x) $ is $ [8, \infty) $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Points of Discontinuity

Points of discontinuity are values of x where the function becomes undefined. These points often occur when the denominator of a fraction is zero, resulting in division by zero. For the function $ r(x) = \frac{1}{x^2} + \frac{1}{(x-1)^2} $, we need to find when the denominators $ x^2 $ and $ (x-1)^2 $ are zero.

To do this, solve the equations:

$ x^2 = 0 \Rightarrow x = 0 $
$ (x-1)^2 = 0 \Rightarrow x = 1 $

Therefore, the function is undefined at $ x = 0 $ and $ x = 1 $, meaning these are points of discontinuity. Recognizing these points helps us understand where the function breaks or has gaps.

Function Behavior

Understanding the behavior of a function near points of discontinuity is crucial. For $ r(x) $, as $ x $ approaches 0 or 1, each term ($ \frac{1}{x^2} $ and $ \frac{1}{(x-1)^2} $) becomes very large (tending towards infinity).

Near the discontinuities:

When $ x $ is close to 0 from either side, $ \frac{1}{x^2} $ grows extremely large, causing $ r(x) $ to increase significantly.
Similarly, approaching 1 from either side causes $ \frac{1}{(x-1)^2} $ to blow up, making $ r(x) $ very large.

This leads to vertical asymptotes at $ x = 0 $ and $ x = 1 $.

Away from these points, the function remains positive and decreases to a minimum value at $ x = 0.5 $, where $ r(0.5) = 8 $. The function never becomes negative or reaches zero, staying positive throughout its domain.

Interval Notation

Interval notation is a way to describe the domain and range of a function. It uses parentheses $( )$ for values that are not included and brackets $[ ]$ for values that are included.

For the domain of $ r(x) $, excluding $ x = 0 $ and $ x = 1 $, we write it in interval notation as:

$ (-\infty, 0) \cup (0, 1) \cup (1, \infty) $

This indicates all real numbers except 0 and 1.

For the range, since the function is always positive and its minimum value is 8, the range in interval notation is:

$ [8, \infty) $

This shows that $ r(x) $ takes on all values from 8 upwards, extending to infinity.

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91影视

Determine the domain and estimate the range of each function. $$ r(x)=\frac{1}{x^{2}}+\frac{1}{(x-1)^{2}} $$

Short Answer

Step by step solution

- Determine Points of Discontinuity

- Solve for Undefined Points

- State the Domain

- Analyze the Behavior at Discontinuities

- Determine the Minimum Value

- State the Range

Key Concepts

Points of Discontinuity

Function Behavior

Interval Notation

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