Problem 279 - Calculus Volume 1

Question

Find the horizontal and vertical asymptotes. \(f(x)=\frac{1}{x^{3}+x^{2}}\)

Step-by-Step Solution

Answer

Vertical asymptotes at \(x = 0\) and \(x = -1\); horizontal asymptote at \(y = 0\).

Step 1: Finding Vertical Asymptotes

Vertical asymptotes occur where the denominator of the function equals zero. Set the denominator equal to zero and solve for the values of \(x\): \(x^3 + x^2 = 0\). Factor the equation as \(x^2(x+1) = 0\). Thus, \(x = 0\) and \(x = -1\) are the points where the denominator is zero, indicating potential vertical asymptotes.

Step 2: Checking for Vertical Asymptotes Points

Verify that the values \(x = 0\) and \(x = -1\) do not result in removable discontinuities. For removable discontinuities, both the numerator and denominator have a common factor. In \(f(x) = \frac{1}{x^3 + x^2}\), the numerator is constant and does not share any factors with the denominator terms \(x^2\) or \(x+1\). Therefore, \(x = 0\) and \(x = -1\) are indeed vertical asymptotes.

Step 3: Finding Horizontal Asymptotes

Horizontal asymptotes occur by comparing the degrees of the numerator and denominator. Here, the numerator is of degree 0 (constant) and the denominator is of degree 3. Since the degree of the numerator (0) is less than the degree of the denominator (3), the horizontal asymptote is \(y = 0\).

Key Concepts

Vertical Asymptotes

Vertical asymptotes are lines that show where a function approaches infinity or negative infinity as the input (usually represented by \(x\)) gets very close to a certain value. They occur where the denominator of a rational function is zero, which makes the function undefined at those points.
For the function \(f(x)=\frac{1}{x^{3}+x^{2}}\), we look at the denominator, \(x^3 + x^2\).
To find potential vertical asymptotes, set the denominator equal to zero:
\[x^3 + x^2 = 0\]Here, we can factor the denominator as:

• First, factor out the common term \(x^2\): \(x^2(x+1) = 0\)
• This gives us solutions \(x = 0\) and \(x = -1\)

At these solutions, the function \(f(x)\) becomes undefined, indicating where vertical asymptotes occur.
It's crucial to check if these points are indeed vertical asymptotes by ensuring they are not removable discontinuities.

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of a graph as \(x\) approaches infinity or negative infinity.
They show the value that the function approaches in the long run, helping us understand end-behavior.
For a rational function \(\frac{a(x)}{b(x)}\), the horizontal asymptotes are determined by comparing the leading degrees of the numerator and the denominator.

• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\).
• If the degrees are the same, the horizontal asymptote is \(y = \frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}\).
• If the degree of the numerator is greater, there are no horizontal asymptotes, although there may be an oblique asymptote.

For our function \(f(x)=\frac{1}{x^{3}+x^{2}}\), the degree of the numerator is 0, and the degree of the denominator is 3.
Therefore, the horizontal asymptote is \(y=0\) because the numerator’s degree is less than the denominator’s.

Removable Discontinuities

Removable discontinuities occur in functions where a point is undefined or not simplified further, but by factoring and simplifying, the discontinuity can be eliminated.
This typically happens when both the numerator and the denominator share a common factor.
In our function \(f(x)=\frac{1}{x^{3}+x^{2}}\), the numerator is a constant 1 and doesn’t contain any factors to cancel with the denominator term \(x^3 + x^2\).

• Because there are no common factors, the denominator cannot be simplified to eliminate the discontinuity.

Therefore, the points \(x = 0\) and \(x = -1\) remain as vertical asymptotes rather than becoming removable discontinuities.
Simply put, a removable discontinuity could be imagined as a 'hole' in the graph that can be patched by simply redefining the function at that point, which isn’t possible here.