Problem 2
Question
Find any points of discontinuity for each rational function. $$ y=\frac{x^{2}+2 x}{x^{2}+2} $$
Step-by-Step Solution
Verified Answer
The function \(y=\frac{x^{2}+2 x}{x^{2}+2}\) has no points of discontinuity.
1Step 1: Address the Denominator
Firstly, set the denominator equal to zero and solve for the variable \(x\): \(x^{2} + 2 = 0\). After subtracting 2 from both sides, the equation becomes \(x^{2} = -2\). Since there's no real number that can be squared to produce a negative result, there are no real roots for this equation. Hence, the denominator will never be zero for any real numbers.
2Step 2: Discuss the Result
Given that the denominator does not equals zero for any real numbers, there are no points of discontinuity for the rational function. This suggests that the function is continuous for all real numbers.
Key Concepts
Points of DiscontinuityDenominator in Rational FunctionsContinuous Functions
Points of Discontinuity
In mathematics, especially in calculus, identifying points of discontinuity in a rational function is crucial. A point of discontinuity, simply put, is a value of the variable where the function is not defined. For rational functions, these points usually occur where the denominator is zero. However, if simplifying the expression or other manipulations remove these potential points, they may not lead to discontinuities in the graph of the function.
- When considering the given function \( y = \frac{x^{2}+2x}{x^{2}+2} \), we analyze the denominator \( x^{2} + 2 \).
- This also involves checking if setting this expression to zero and solving for \( x \) yields any real solutions.
Denominator in Rational Functions
In rational functions, the denominator plays a pivotal role in determining the function's behavior. A rational function is of the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. The denominator \( Q(x) \) is essential because its value can define or disrupt the function's continuity.
Since this equation \( x^{2} = -2 \) does not have real solutions, the denominator is never zero. Understanding this property of the denominator helps us conclude that there are no discontinuities for real numbers in this function.
- If the denominator becomes zero, the function is undefined at that point.
- Resolving \( Q(x) = 0 \) can help identify where these potential problems might exist.
Since this equation \( x^{2} = -2 \) does not have real solutions, the denominator is never zero. Understanding this property of the denominator helps us conclude that there are no discontinuities for real numbers in this function.
Continuous Functions
A function is continuous if its graph can be drawn without lifting the pencil from the paper. In mathematical terms, a function is continuous at a point if it is defined there, the limit as it approaches that point exists, and the value of the function equals the value of the limit.
For rational functions, continuity across their domains is defined by the absence of points where the function becomes undefined; that is when the denominator is zero.
For rational functions, continuity across their domains is defined by the absence of points where the function becomes undefined; that is when the denominator is zero.
- Our function \( y = \frac{x^{2}+2x}{x^{2}+2} \) illustrates this case perfectly.
- Because the denominator \( x^{2} + 2 \) never equals zero for any real number, the function remains continuous along the entire real number line.
- Therefore, there are no gaps, jumps, or breaks within this function.
Other exercises in this chapter
Problem 2
Solve each equation. Check each solution. $$ \frac{1}{5 x}=\frac{1}{9 x} $$
View solution Problem 2
Simplify each rational expression. State any restrictions on the variable. $$ \frac{6 c^{2}+9 c}{3 c} $$
View solution Problem 2
Make a table of values. Then sketch a graph of each inverse variation. \(y=\frac{10}{x}\)
View solution Problem 2
Suppose that \(x\) and \(y\) vary inversely. Write a function that models each inverse variation. $$ x=-13 \text { when } y=100 $$
View solution