Problem 260 - Calculus Volume 1

Question

Determine whether there is an asymptote at \(x=a\). Justify your answer without graphing on a calculator. \(f(x)=1+x^{-2 / 5}, a=1\)

Step-by-Step Solution

Answer

There is no asymptote at \(x=1\) since the function is defined and finite there.

Step 1: Understand Asymptotes

An asymptote is a line that a graph approaches but never actually reaches. For vertical asymptotes, we look for values of \(x\) that make the function undefined.

Step 2: Evaluate the Function at x=a

Substitute \(x=1\) into the function \(f(x) = 1 + x^{-2/5}\). This becomes \(f(1) = 1 + 1^{-2/5} = 1 + 1 = 2\). This result means the function is defined and finite at \(x=1\).

Step 3: Check for Vertical Asymptotes

Since a vertical asymptote occurs when a function is undefined (usually due to division by zero), and \(f(x)\) is defined and finite at \(x=1\), there is no vertical asymptote at \(x=1\).

Step 4: Conclusion on Asymptotes

Since the function \(f(x)=1+x^{-2/5}\) is defined at \(x=1\), there is no asymptote of any kind at \(x=1\).

Key Concepts

Vertical Asymptotes

Vertical asymptotes occur in a graph when the function approaches a certain line but never quite touches it. This happens when you try to divide by zero or encounter an undefined expression. For example, if you have a fraction where the denominator goes to zero as \( x \) approaches a certain value, that's when a vertical asymptote is likely present.

• These asymptotes manifest as vertical lines on a graph.
• They indicate where the function shoots up to positive or negative infinity.

To check if there's a vertical asymptote at \( x = a \), substitute \( x \) with \( a \) in your function. If the function becomes undefined, that likely points to a vertical asymptote. In our exercise, though, evaluating \( f(1) = 1 + 1^{-2/5} \) gives 2, a perfectly defined result, ruling out the presence of a vertical asymptote at \( x = 1 \).

Function Analysis

Function analysis involves examining a function's behavior and properties with various techniques. To perform a thorough function analysis, you typically:

• Identify any asymptotes, intercepts, and zeros.
• Evaluate the limits to understand the end behavior of the function.

The function \( f(x) = 1 + x^{-2/5} \) doesn't have vertical asymptotes at \( x = 1 \) because it results in a finite value. However, beyond asymptotes, function analysis encompasses a broader examination:

• Check if the function is continuous across the domain.
• Look at derivatives to find increasing or decreasing intervals.
• Determine concavity and inflection points.

Doing so gives a clearer picture of the full behavior of the function.

Undefined Functions

Functions become undefined in certain conditions, typically when you have:

• Division by zero, which makes a function cease to deliver real numbers.
• Logarithms of zero or negative numbers.
• Square roots of negative numbers when dealing with real functions.

In our specific exercise, checking for undefined functions involves substituting \( x = 1 \) into \( f(x) \) to ensure no division by zero or other undefined operations occur.
With \( f(x) = 1 + x^{-2/5} \), substituting gives \( 2 \), a definite and finite number. Understanding why a function may be undefined ensures clarity on whether asymptotes or other discontinuities are present.