Problem 13
Question
Describe the interval(s) on which the function is continuous. Explain why the function is continuous on the interval(s). If the function has a discontinuity, identify the conditions of continuity that are not satisfied. \(f(x)=\frac{x^{2}-1}{x+1}\)
Step-by-Step Solution
Verified Answer
The function \(f(x)=\frac{x^{2}-1}{x+1}\) is continuous for all real values except at \(x=-1\). At \(x=-1\), the function is undefined and thus has a discontinuity.
1Step 1: Identifying the point of discontinuity
Since \(f(x)=\frac{x^{2}-1}{x+1}\), the function is undefined and thus discontinuous at \(x=-1\) because the denominator becomes zero at this point. This has to be checked first. If we plug \(x=-1\) into the function, we get that the function is undefined.
2Step 2: Simplifying the equation
Simplify \(f(x)\) by factoring the numerator. The function becomes \(f(x) = \frac{(x-1)(x+1)}{x+1}\). If we cancel out the common factor \(x+1\) from numerator and denominator, the function becomes \(f(x)=x-1\). But remember, this is true only for \(x≠-1\). We cannot forget about that point where the original function was undefined.
3Step 3: Checking the conditions of continuity
The function \(f(x)=x-1\) is a linear function and its domain is all real numbers. Thus, it is continuous for all \(x≠-1\). But at \(x=-1\), the original function was undefined, none of the continuity conditions are satisfied. Therefore, there is a discontinuity at \(x=-1\).
Key Concepts
Point of DiscontinuitySimplifying Rational ExpressionsConditions of Continuity
Point of Discontinuity
When exploring functions for continuity, identifying points of discontinuity is crucial. A point of discontinuity occurs where a function is not defined or shifts behavior. For the function \( f(x)=\frac{x^{2}-1}{x+1} \), such a point is found where the denominator is zero.
In this function, plugging in \( x = -1 \) makes the denominator zero, rendering the function undefined. Hence, \( x = -1 \) is the point of discontinuity. Pinpointing these spots helps in understanding how the function behaves at or near these values.
Key points to remember about discontinuity include:
In this function, plugging in \( x = -1 \) makes the denominator zero, rendering the function undefined. Hence, \( x = -1 \) is the point of discontinuity. Pinpointing these spots helps in understanding how the function behaves at or near these values.
Key points to remember about discontinuity include:
- The function is undefined at this point.
- Mathematical operations like division by zero often lead to discontinuities.
- These points are essential when evaluating limits and continuity conditions.
Simplifying Rational Expressions
Rational expressions, like \( \frac{x^2-1}{x+1} \), often need simplification to better understand their properties. This involves simplifying the numerator and the denominator when possible.
For the function given, the numerator \( x^2-1 \) can be factored into \( (x-1)(x+1) \). This leads to \( f(x) = \frac{(x-1)(x+1)}{x+1} \).
By simplifying, we remove the common factor \( x+1 \) from both numerator and denominator, resulting in \( f(x) = x-1 \) for \( x eq -1 \).
When dealing with rational expressions, keep in mind:
Thus, understanding how to simplify rational expressions can make the analysis of a function's continuity clearer.
For the function given, the numerator \( x^2-1 \) can be factored into \( (x-1)(x+1) \). This leads to \( f(x) = \frac{(x-1)(x+1)}{x+1} \).
By simplifying, we remove the common factor \( x+1 \) from both numerator and denominator, resulting in \( f(x) = x-1 \) for \( x eq -1 \).
When dealing with rational expressions, keep in mind:
- Factor common terms where possible.
- A simplified expression might reveal more about a function's behavior.
- Simplification doesn’t change the undefined points of the original expression.
Thus, understanding how to simplify rational expressions can make the analysis of a function's continuity clearer.
Conditions of Continuity
The conditions of continuity dictate whether a function is continuous over an interval. For a function to be continuous at a point \( c \), three conditions must be met:
For our function, \( f(x)=x-1 \), these conditions hold true for all \( x eq -1 \). However, at \( x = -1 \), the original function \( \frac{x^2-1}{x+1} \) is not defined, failing the first condition.
Understanding these conditions helps in identifying where a function is continuous and where its behavior changes significantly. Points of discontinuity, like \( x = -1 \), are places where these conditions break down.
Continuity is key in calculus as it allows us to apply many important theorems and tools.
- The function \( f(x) \) must be defined at \( c \).
- The limit of \( f(x) \) as \( x \) approaches \( c \) must exist.
- The limit of \( f(x) \) as \( x \) approaches \( c \) must equal \( f(c) \).
For our function, \( f(x)=x-1 \), these conditions hold true for all \( x eq -1 \). However, at \( x = -1 \), the original function \( \frac{x^2-1}{x+1} \) is not defined, failing the first condition.
Understanding these conditions helps in identifying where a function is continuous and where its behavior changes significantly. Points of discontinuity, like \( x = -1 \), are places where these conditions break down.
Continuity is key in calculus as it allows us to apply many important theorems and tools.
Other exercises in this chapter
Problem 13
Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$ g(x)=\frac{2 x+1}{x-5}
View solution Problem 13
Find the derivative of the function. $$ f(t)=-3 t^{2}+2 t-4 $$
View solution Problem 13
Find the limit of (a) \(f(x)+g(x)\), (b) \(f(x) g(x)\), and (c) \(f(x) / g(x)\), as \(x\) approaches \(c\). $$ \begin{aligned} &\lim _{x \rightarrow c} f(x)=3 \
View solution Problem 14
Find \(d y / d u, d u / d x\), and \(d y / d x\). $$ y=u^{-1}, u=x^{3}+2 x^{2} $$
View solution