Problem 204 - Precalculus

Question

For the following exercises, state the domain, vertical asymptote, and end behavior of the function. $$f(x)=\ln (2-x)$$

Step-by-Step Solution

Answer

Domain: \((-\infty, 2)\), Vertical Asymptote: \(x = 2\), End behavior: \(x \to -\infty, f(x) \to \infty\); \(x \to 2^-, f(x) \to -\infty\).

Step 1: Determine the Domain

The function is a logarithmic function, \(f(x) = \ln (2-x)\). The domain of a logarithmic function is determined by the logarithm argument being greater than zero. Therefore, we set \(2-x > 0\) and solve for \(x\).\[2 - x > 0\]\[x < 2\]The domain of the function is all \(x\) such that \(x < 2\), which in interval notation is \((-\infty, 2)\).

Step 2: Identify the Vertical Asymptote

The vertical asymptote of a logarithmic function occurs where the argument of the logarithm is zero. This can be found by setting \(2-x = 0\) and solving for \(x\).\[2 - x = 0\]\[x = 2\]Thus, there is a vertical asymptote at \(x = 2\).

Step 3: Analyze End Behavior as \(x \to -\infty\)

For large negative values of \(x\), \(2-x\) becomes large and positive, making \(\ln(2-x)\) approach \(\ln(\infty)\), which is also infinite. As \(x \to -\infty\), \(f(x) \to \infty\).

Step 4: Analyze End Behavior as \(x \to 2^-\)

When \(x\) approaches 2 from the left (\(x \to 2^-\)), \(2-x\) approaches zero from the positive side. In this case, \(\ln(2-x)\) tends towards \(-\infty\). Thus, as \(x \to 2^-\), \(f(x) \to -\infty\).

Key Concepts

Domain

The domain of a function is a set of all possible input values (usually represented as \(x\)) for which the function is defined. When dealing with logarithmic functions like \(f(x) = \ln(2-x)\), the domain is determined by the argument of the logarithm. For a logarithmic function \(\ln(a)\), the argument \(a\) must be greater than zero. Therefore, for \(\ln(2-x)\), we must ensure that \(2-x > 0\). Solving this, we get:

• \(2 - x > 0\)
• \(x < 2\)

This inequality tells us that \(x\) can be any value less than 2. In interval notation, this is expressed as \(( -\infty, 2)\). This means the function can accept any real number into it, as long as it is less than 2. By understanding these input limits, you ensure that the function remains real and defined.

Vertical Asymptotes

Vertical asymptotes are vertical lines that the graph of a function approaches but never touches or crosses. They occur at points where the function increases or decreases without bound. For the function \(f(x) = \ln(2-x)\), vertical asymptotes are found by setting the argument of the logarithm to zero, This requires that the equation:

• \(2-x = 0\)

be solved. Solving it gives: \(x = 2\). Thus, the line \(x = 2\) is a vertical asymptote. As \(x\) gets really close to 2 from the left side, the function \(f(x)\) itself tends towards negative infinity, indicating it does not actually reach the asymptote value but goes down infinitely. Vertical asymptotes help us understand the behavior of functions near certain critical values and are essential in comprehensively analyzing the function's graph.

End Behavior

End behavior in functions describes what the function's output or \(y\)-values are doing as the input or \(x\)-values move towards positive or negative infinity. For \(f(x) = \ln(2-x)\), analyzing end behavior involves looking at two scenarios.Firstly, consider when \(x\) moves towards \(-\infty\). In this case, \(2-x\) becomes a very large positive number. The logarithm of a large positive number grows indefinitely large, so \(\ln(2-x)\) approaches \(\infty\). Therefore, as \(x \to -\infty\), \(f(x) \to \infty\).Next, when \(x\) approaches the asymptote at \(x = 2\) from the left, noted as \(x \to 2^-\), \(2-x\) approaches zero from the positive side. As a result, \(\ln(2-x)\) approaches \(-\infty\). This means that as \(x\) gets closer to 2 from the left, \(f(x)\) descends towards \(-\infty\). Understanding end behavior gives insights into how the function reacts at extreme values of \(x\), informing us about its long-term behavior on a graph.