Problem 35 - Algebra and Trigonometry with Analytic Geometry

Question

Sketch the graph of \(f\). $$ f(x)=\log _{3}(3 x) $$

Step-by-Step Solution

Answer

The graph of \( f(x) \) is \( 1 + \log_3(x) \), shifted up by 1 unit with a vertical asymptote at \( x = 0 \).

Step 1: Understand the Function

The function given is \( f(x) = \log_3(3x) \). This is a logarithmic function with base 3 and argument \(3x\). In logarithmic functions, the expression inside the log determines its domain and behavior. This function is defined for \(3x > 0\), which simplifies to \(x > 0\). The function will not be defined for \(x \leq 0\).

Step 2: Simplify the Function

We can simplify the function \( f(x) = \log_3(3x) \) by using the logarithmic identity \( \log_b(a \cdot c) = \log_b(a) + \log_b(c) \). Here, \( a = 3 \) and \( c = x \), so \( f(x) = \log_3(3) + \log_3(x) \). Knowing that \( \log_3(3) = 1 \), we have \( f(x) = 1 + \log_3(x) \).

Step 3: Identify Key Features

Since \( f(x) = 1 + \log_3(x) \), it's helpful to understand how a standard logarithmic function \( \log_3(x) \) behaves. It has a vertical asymptote at \( x = 0 \), passes through the point \( (1, 0) \), and slowly increases to the right without bound. Adding 1 translates the graph up by 1 unit.

Step 4: Graph the Function

Start by plotting the basic shape of \( \log_3(x) \), which is a curve starting near negative infinity as \( x \to 0^+ \) and gradually increasing. Shift the entire curve up by 1 unit to reflect \( f(x) = \log_3(x) + 1 \). Now, it passes through \( (1,1) \) instead of \( (1,0) \). Remember the asymptote at \( x = 0 \).

Step 5: Analyze Behavior at Intervals

For \( x > 1 \), the function \( f(x) \) continues increasing because \( \log_3(x) \) increases. For \( 0 < x < 1 \), \( \log_3(x) \) is negative, but \( f(x) = 1 + \log_3(x) \) remains below 1 yet above -∞. There is no horizontal asymptote; the function will increase constantly to the right.

Key Concepts

Domain and Range

When exploring logarithmic functions like \(f(x) = \log_3(3x)\), it's essential to understand their domain and range. The domain refers to all the possible input values \(x\) can take for the function to be valid. Since the logarithm is undefined for negative numbers or zero, its argument must be positive. Thus, for \(f(x) = \log_3(3x)\), we set \(3x > 0\). Simplifying this inequality, we find that \(x > 0\). Therefore, the domain of \(f(x)\) is all positive numbers, \(x \in (0, \infty)\).

The range of a logarithmic function includes all real numbers. Once you simplify the function to \(f(x) = 1 + \log_3(x)\), the transformation shifts each output of \(\log_3(x)\) up by 1. The function can still output any real number, so its range remains \((-\infty, \infty)\). Understanding these properties is crucial when graphing or analyzing the behavior of the function.

Vertical Asymptotes

Vertical asymptotes represent points where a function approaches infinity or negative infinity. They signify boundaries that the function cannot cross. For the function \(f(x) = 1 + \log_3(x)\), there is a vertical asymptote at \(x = 0\). This occurs because as \(x\) approaches zero from the right (\(x \to 0^+\)), the logarithmic part \(\log_3(x)\) approaches negative infinity. Consequently, \(f(x)\) also approaches negative infinity.

• The graph of \(f(x)\) will get closer and closer to the vertical line \(x = 0\) but can never actually reach or cross it.
• This is true for any logarithmic function with an argument \(x\) that must remain positive for valid computation.
• Vertical asymptotes are common in logarithmic graphs and serve as indicators of their input restrictions.

Knowing this helps prevent confusion with graph behavior near \(x = 0\).

Transformations of Functions

Transformations allow us to modify the basic graph of a function in predictable ways. When simplifying \(f(x) = \log_3(3x)\) into \(f(x) = 1 + \log_3(x)\), we recognize a specific vertical shift applied to the base function \(\log_3(x)\). This shift occurs because adding 1 moves the entire graph one unit upwards.

• A vertical shift by 1 unit up means that every point on \(\log_3(x)\) now has its \(y\)-value increased by 1.
• This transformation affects specific points; for example, \((1, 0)\) on \(\log_3(x)\) becomes \((1, 1)\) on \(f(x)\).
• The rise in the graph does not affect the asymptote at \(x = 0\), as vertical transformations don't alter asymptotic behavior.

Transformations like this help us comfortably adjust basic functions to new forms while maintaining understanding of their shifts and changes.