Problem 76
Question
For the following exercises, identify the removable discontinuity. $$ f(x)=\frac{x^{3}+1}{x+1} $$
Step-by-Step Solution
Verified Answer
The removable discontinuity is at \(x = -1\).
1Step 1: Identify Points of Discontinuity
To identify points of discontinuity, find the values of \(x\) for which the denominator of the function is equal to zero. Set \(x + 1 = 0\) and solve for \(x\). This gives \(x = -1\) as the point where the function could be discontinuous.
2Step 2: Simplify the Function
Factor the numerator \(x^{3} + 1\) using the sum of cubes formula: \(a^{3} + b^{3} = (a + b)(a^2 - ab + b^2)\). Here, \(a = x\) and \(b = 1\), so: \[ x^{3} + 1 = (x + 1)(x^2 - x + 1) \].
3Step 3: Cancel Common Factors
The expression \(\frac{x^{3} + 1}{x + 1}\) becomes \(\frac{(x + 1)(x^2 - x + 1)}{x + 1}\). Cancel the common factor \(x + 1\) in the numerator and denominator, leading to the simplified form \(x^2 - x + 1\).
4Step 4: Confirm Removable Discontinuity
Since the discontinuity at \(x = -1\) is removed by canceling the \(x + 1\) term, this point is a removable discontinuity. The function can be redefined at \(x = -1\) using the simplified function \(f(x) = x^2 - x + 1\).
Key Concepts
Points of DiscontinuitySum of CubesSimplifying Rational ExpressionsAlgebraic Functions
Points of Discontinuity
A point of discontinuity in a function occurs where the function is not defined. These points cause the function to "break" at certain values of x. To locate them, check the denominator of the rational expression for any values that make it zero, as division by zero is undefined in mathematics.
In our example, the function is \( f(x) = \frac{x^3 + 1}{x + 1} \). Set the denominator \( x + 1 = 0 \). Solving this equation gives \( x = -1 \). This means \( x = -1 \) is a point where the function could break or become discontinuous. However, this doesn't always mean the discontinuity is significant. It could be removable, as we'll see in later sections.
In our example, the function is \( f(x) = \frac{x^3 + 1}{x + 1} \). Set the denominator \( x + 1 = 0 \). Solving this equation gives \( x = -1 \). This means \( x = -1 \) is a point where the function could break or become discontinuous. However, this doesn't always mean the discontinuity is significant. It could be removable, as we'll see in later sections.
Sum of Cubes
The numerator of our function \( x^3 + 1 \) is a classic example of a sum of cubes. The sum of cubes formula is: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). This is essential for breaking down algebraic expressions into more manageable parts.
For this function, recognize it as \( a^3 + 1^3 \), where \( a = x \) and \( b = 1 \). Using the formula, we can express \( x^3 + 1 = (x + 1)(x^2 - x + 1) \).
For this function, recognize it as \( a^3 + 1^3 \), where \( a = x \) and \( b = 1 \). Using the formula, we can express \( x^3 + 1 = (x + 1)(x^2 - x + 1) \).
- This factorization helps by revealing any common factors to be simplified later.
- Used in simplifying rational expressions, it shows hidden structures within polynomials.
Simplifying Rational Expressions
Simplifying rational expressions involves techniques like factoring common terms and canceling them. This process especially helps in identifying removable discontinuities.
In our function \( \frac{x^3 + 1}{x + 1} \), after factoring the numerator, we can express it as \( \frac{(x + 1)(x^2 - x + 1)}{x + 1} \). Now, you can see \( (x + 1) \) appears in both numerator and denominator. Cancel these common terms, leading to the simplified form \( x^2 - x + 1 \).
In our function \( \frac{x^3 + 1}{x + 1} \), after factoring the numerator, we can express it as \( \frac{(x + 1)(x^2 - x + 1)}{x + 1} \). Now, you can see \( (x + 1) \) appears in both numerator and denominator. Cancel these common terms, leading to the simplified form \( x^2 - x + 1 \).
- Canceling common factors rectifies the discontinuity at \( x = -1 \), making it removable.
- This step not only simplifies the expression but clarifies the nature of the discontinuity.
Algebraic Functions
Algebraic functions like \( f(x) = \frac{x^3 + 1}{x + 1} \) consist of quotients of polynomial expressions. They can represent complex behaviors such as asymptotes and discontinuities.
Algebraic functions can exhibit different kinds of discontinuities: removable, jump, and infinite. Our focus here is on removable discontinuities, where the discontinuity can be "repaired" or "removed" by simplifying the function.
Once the \( x + 1 \) is canceled, the function \( f(x) = x^2 - x + 1 \) becomes well-defined even at \( x = -1 \), except for the minor tweak needed to redefine it at \( x = -1 \).
Algebraic functions can exhibit different kinds of discontinuities: removable, jump, and infinite. Our focus here is on removable discontinuities, where the discontinuity can be "repaired" or "removed" by simplifying the function.
Once the \( x + 1 \) is canceled, the function \( f(x) = x^2 - x + 1 \) becomes well-defined even at \( x = -1 \), except for the minor tweak needed to redefine it at \( x = -1 \).
- Redefining ensures continuity throughout the function's domain.
- Understanding these dynamics of algebraic functions aids in proper interpretation and manipulation of their graphs.
Other exercises in this chapter
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