Problem 20 - College Algebra

Question

For the following exercises, state the domain, vertical asymptote, and end behavior of the function. $$f(x)=\log _{3}(15-5 x)+6$$

Step-by-Step Solution

Answer

Domain: \((-\infty, 3)\); Vertical asymptote: \(x = 3\); End behavior: As \(x \to -\infty, f(x) \to \infty\) and \(x \to 3^-, f(x) \to -\infty\).

Step 1: Determine the domain

The domain of a logarithmic function is determined by the argument inside the logarithm. Since \(f(x) = \log_{3}(15 - 5x) + 6\), the argument is \(15 - 5x\). The argument must be greater than zero: \(15 - 5x > 0\). Solving this inequality: \[15 > 5x \] \[\frac{15}{5} > x \] \[3 > x\] or \(x < 3\). Therefore, the domain is \((-\infty, 3)\).

Step 2: Identify the vertical asymptote

The vertical asymptote occurs where the logarithm is undefined, which is when the argument equals zero. Set the argument of the logarithm equal to zero: \(15 - 5x = 0\). Solving for \(x\): \[15 = 5x\] \[x = 3\]. Thus, the vertical asymptote is \(x = 3\).

Step 3: Analyze the end behavior

To determine the end behavior, consider the behavior of \(f(x)\) as \(x\) approaches the boundary of its domain. As \(x\) approaches 3 from the left (\(x \to 3^-\)), \(15 - 5x\) approaches 0 from the right. The logarithmic function approaches negative infinity in this case because \(\log_{3}(0^+) = -\infty\). As \(x\) approaches negative infinity, \(15 - 5x\) approaches a large positive value, so \(\log_{3}(15 - 5x)\) approaches infinity, resulting in \(f(x) \to \infty\). Therefore, as \(x \to -\infty\), \(f(x) \to \infty\) and as \(x \to 3^-\), \(f(x) \to -\infty\).

Key Concepts

Domain of a Function

The domain of a function refers to all the possible input values (x-values) that the function can accept without causing any mathematical inconsistencies. In the context of our logarithmic function, it is crucial to ensure that the argument inside the logarithm is positive, because the logarithm of zero or a negative number is undefined. For our function, the argument is \(15 - 5x\). We need to solve the inequality \(15 - 5x > 0\) to determine which x-values are acceptable.

Breaking it down:

• Start by isolating \(x\) in the inequality \(15 - 5x > 0\).
• Rearrange to find \(15 > 5x\) by moving \(-5x\) to the right side.
• Divide both sides by 5 to get \(3 > x\), or equivalently, \(x < 3\).

Hence, the domain of our function is all values of \(x\) that are less than 3, expressed in interval notation as \((-\infty, 3)\).

Vertical Asymptotes

Vertical asymptotes represent the values of \(x\) where a function's value tends to infinity, positive or negative, indicating a break or undefined point in the curve. For logarithmic functions, a vertical asymptote typically appears where the function's argument equals zero. In our function, this occurs when \(15 - 5x = 0\).

To find this:

• Solve the equation \(15 - 5x = 0\).
• Add \(5x\) to both sides yielding \(15 = 5x\).
• Then, divide both sides by 5 to solve for \(x\), resulting in \(x = 3\).

Therefore, our function has a vertical asymptote at \(x = 3\), meaning the function increases or decreases without bounds as \(x\) approaches 3 from either direction.

End Behavior Analysis

End behavior analysis involves understanding how a function behaves as the input values move towards the boundaries of the function's domain. For the logarithmic function \(f(x) = \log_{3}(15 - 5x) + 6\), it is important to consider what happens as \(x\) nears the upper boundary of the domain, and as it approaches negative infinity.

To analyze the end behavior:

• As \(x\) gets closer to 3 from the left (\(x \to 3^-\)), \(15 - 5x\) closes in on zero from the positive side. Hence, \(\log_{3}(15 - 5x)\) approaches \(-\infty\), leading \(f(x)\) to also tend towards \(-\infty\).
• Conversely, as \(x\) approaches \(-\infty\), the argument \(15 - 5x\) becomes a large positive number, making \(\log_{3}(15 - 5x)\) approach \(\infty\). Thus, \(f(x)\) heads towards \(\infty\).

In summary, as \(x\) approaches the left-side boundary 3, our function tends towards \(-\infty\), and as \(x\) stretches to negative infinity, it increases infinitely.