Problem 17
Question
In Problems \(15-24\), find the values of \(x \in \mathbf{R}\) for which the given functions are continuous. $$ f(x)=\frac{x^{2}+1}{x-1} $$
Step-by-Step Solution
Verified Answer
The function is continuous for \( x \in (-\infty, 1) \cup (1, \infty) \).
1Step 1: Identify Potential Discontinuities
A function is potentially discontinuous where the denominator is zero. For the function \( f(x) = \frac{x^2 + 1}{x - 1} \), we see that the denominator is \( x - 1 \). Set this equal to zero to find where \( f(x) \) is undefined: \( x - 1 = 0 \). Solving this gives \( x = 1 \).
2Step 2: Analyze Continuity at the Critical Point
For a rational function like \( f(x) = \frac{x^2 + 1}{x - 1} \), continuity is present everywhere except where the denominator is zero. We found in Step 1 that the function is undefined at \( x = 1 \), indicating the function is discontinuous at this point.
3Step 3: State the Interval of Continuity
The function \( f(x) = \frac{x^2 + 1}{x - 1} \) is continuous for all real numbers except \( x = 1 \). Therefore, the interval of continuity is \((-\infty, 1) \cup (1, \infty)\).
Key Concepts
DiscontinuityRational FunctionsInterval of Continuity
Discontinuity
Understanding discontinuity is crucial when studying functions, especially rational ones. A discontinuity is a point where a function is not continuous, which generally means the function has a break or a gap at that point. For most functions, discontinuity occurs where the function is undefined. With rational functions, this often happens when the denominator equals zero.
For example, consider the function \( f(x) = \frac{x^2 + 1}{x - 1} \). The denominator, \(x - 1\), becomes zero when \(x = 1\). At this specific value of \(x\), the function is undefined, resulting in a discontinuity. Therefore, \(x = 1\) is a point of discontinuity for \(f(x)\).
This understanding helps in determining where the function "breaks" in its behavior. Identifying such points is the first step when solving problems related to function continuity.
For example, consider the function \( f(x) = \frac{x^2 + 1}{x - 1} \). The denominator, \(x - 1\), becomes zero when \(x = 1\). At this specific value of \(x\), the function is undefined, resulting in a discontinuity. Therefore, \(x = 1\) is a point of discontinuity for \(f(x)\).
This understanding helps in determining where the function "breaks" in its behavior. Identifying such points is the first step when solving problems related to function continuity.
Rational Functions
Rational functions are a type of function expressed as the ratio of two polynomials. They have the form \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) eq 0\). These functions can model various behaviors in mathematics due to their versatility.
For \( f(x) = \frac{x^2 + 1}{x - 1} \), the numerator is the polynomial \(x^2 + 1\), and the denominator is \(x - 1\). The behavior of rational functions is significantly influenced by their denominators. When the denominator is zero, the function value tends towards infinity, indicating a vertical asymptote and a discontinuity.
Understanding rational functions helps in analyzing other related features such as horizontal asymptotes, end behavior, and the general shape of the graph. They play an essential role in calculus and algebra, providing insight into real-world applications such as in physics and economics.
For \( f(x) = \frac{x^2 + 1}{x - 1} \), the numerator is the polynomial \(x^2 + 1\), and the denominator is \(x - 1\). The behavior of rational functions is significantly influenced by their denominators. When the denominator is zero, the function value tends towards infinity, indicating a vertical asymptote and a discontinuity.
Understanding rational functions helps in analyzing other related features such as horizontal asymptotes, end behavior, and the general shape of the graph. They play an essential role in calculus and algebra, providing insight into real-world applications such as in physics and economics.
Interval of Continuity
The interval of continuity refers to the set of all values for which a function is continuous. To determine this interval, we first need to identify all points of discontinuity. For rational functions, this involves solving for where the denominator is zero since these are typically the points of discontinuity.
For instance, in the function \( f(x) = \frac{x^2 + 1}{x - 1} \), we identified that \(x = 1\) is a point of discontinuity. Therefore, the function is continuous everywhere else. The interval of continuity can be expressed in interval notation as \(( -\infty, 1 ) \cup ( 1, \infty )\). This notation shows that the function is continuous from negative infinity up to \(x = 1\), and then from just after \(x = 1\) to positive infinity.
Understanding and stating the interval of continuity helps in graphing the function accurately and understanding its behavior over the real numbers. This concept is fundamental in calculus and significantly aids in analyzing the continuity of different types of functions.
For instance, in the function \( f(x) = \frac{x^2 + 1}{x - 1} \), we identified that \(x = 1\) is a point of discontinuity. Therefore, the function is continuous everywhere else. The interval of continuity can be expressed in interval notation as \(( -\infty, 1 ) \cup ( 1, \infty )\). This notation shows that the function is continuous from negative infinity up to \(x = 1\), and then from just after \(x = 1\) to positive infinity.
Understanding and stating the interval of continuity helps in graphing the function accurately and understanding its behavior over the real numbers. This concept is fundamental in calculus and significantly aids in analyzing the continuity of different types of functions.
Other exercises in this chapter
Problem 17
Use the formal definition of limits to prove each statement. $$ \lim _{x \rightarrow \infty} \frac{2}{x^{2}}=0 $$
View solution Problem 17
In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow 0} \frac{1}{1+x^{2}} $$
View solution Problem 17
Evaluate the limits in problems. $$ \lim _{x \rightarrow-\infty} \exp [x] $$
View solution Problem 18
Evaluate the trigonometric limits. $$ \lim _{x \rightarrow 0} \frac{1-\cos (x / 2)}{x} $$
View solution