Problem 20 - College Algebra with Corequisite Support

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

The domain is \((-\infty, 3)\), with a vertical asymptote at \(x = 3\), and as \(x \to 3^{-}\), \(f(x) \to -\infty\); as \(x \to -\infty\), \(f(x) \to \infty\).

Step 1: Find the Domain

The domain of a logarithmic function is determined by the expression inside the logarithm, which must be positive. For the function \(f(x) = \log_{3}(15 - 5x) + 6\), we need \(15 - 5x > 0\). Solve for \(x\): \(15 > 5x\) leads to \(x < 3\). So, the domain is all \(x\) such that \(x < 3\). In interval notation, this is \((-\infty, 3)\).

Step 2: Determine the Vertical Asymptote

The vertical asymptote of a logarithmic function occurs where the expression inside the logarithm is zero. \(15 - 5x = 0\) gives us \(x = 3\). Therefore, there is a vertical asymptote at \(x = 3\).

Step 3: Analyze the End Behavior

To determine the end behavior, observe how \(f(x)\) behaves as \(x\) approaches the boundary of its domain, which is \(x = 3\). As \(x \rightarrow 3^{-}\) (approaching from the left), \(f(x)\) approaches \(-\infty\) because the logarithm of a number that is close to zero but positive becomes very large in the negative direction. For \(x \rightarrow -\infty\), \(15 - 5x\) becomes positive and very large, and so \(\log (15 - 5x)\) becomes large in the positive direction, meaning \(f(x) \rightarrow \infty\).

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values (typically "x" values) that allow the function to exist. For logarithmic functions, this is crucial because the expressions inside the log must be positive. Logarithms are defined only for positive numbers. This means for the function given, \( f(x) = \log_{3}(15 - 5x) + 6 \), we solve the inequality \( 15 - 5x > 0 \). Solving:

• Subtract \(15\) from both sides: \(-5x > -15\)
• Divide by \(-5\): \(x < 3\) (note that dividing or multiplying by a negative flips the inequality sign)

Thus, the domain of \( f(x) \) is all real numbers less than 3. Expressed in interval notation, it's \(( -\infty, 3 )\).

Vertical Asymptote

In logarithmic functions, the vertical asymptote is a vertical line on the graph where the function approaches infinity or negative infinity. It occurs where the expression inside the logarithm is equal to zero because that's where the log function is undefined. For the function \( f(x) = \log_{3}(15 - 5x) + 6 \), set the expression inside the log equal to zero:

• \( 15 - 5x = 0 \)
• Solve for \(x: x = 3\)

At \(x = 3\), the logarithm component, \(\log_{3}(15 - 5x)\), is undefined, causing the graph to have a vertical asymptote there. This means as \(x\) approaches 3 from the left, the graph of \(f(x)\) heads toward \(-\infty\).

End Behavior

End behavior helps us understand what happens to the function's values (\(f(x)\)) as \(x\) approaches the boundaries of its domain. For \( f(x) = \log_{3}(15 - 5x) + 6 \), we assess two scenarios:1. As \(x\) approaches 3 from the left (\(x \rightarrow 3^{-}\)): - The expression \(15 - 5x\) is small and positive. - \(\log_{3}(\text{small positive number})\) becomes a large negative number. - \(f(x)\) heads toward \(-\infty\).2. As \(x\) moves to \(-\infty\): - \(15 - 5x\) becomes very large and positive. - \(\log_{3}(\text{large number})\) becomes very large. - \(f(x)\) moves toward \(+\infty\). - This indicates that to the far left of the vertical asymptote, the graph shoots upwards.

Interval Notation

Interval notation is a concise way of expressing the domain or range of a function. It uses parentheses \((\) and brackets \([\) to indicate intervals on the real number line. In interval notation:- Parentheses \((\) are used for numbers that are not included in the interval. For example, \((-\infty, 3)\) means the interval includes every number less than 3, but not 3 itself.- Brackets \([\) are used for numbers that are included in the interval. For instance, \([2, 5]\) includes all numbers from 2 to 5, including both 2 and 5.For the given function \( f(x) = \log_{3}(15 - 5x) + 6 \), the domain is expressed as \((-\infty, 3)\). This implies that all \(x\) values from negative infinity up to, but not including, 3 are permissible inputs for the function.