Problem 25 - College Algebra

Question

For the following exercises, state the domain, range, and \(x\) -and \(y\) -intercepts, if they do not exist, write DNE. $$h(x)=3 \ln (x)-9$$

Step-by-Step Solution

Answer

Domain: (0, ∞), Range: (-∞, ∞), x-intercept: (e³, 0), y-intercept: DNE.

Step 1: Identify the Function and Its Properties

The given function is a logarithmic function defined as \( h(x) = 3 \ln(x) - 9 \). A logarithmic function \( \ln(x) \) is only defined for \( x > 0 \). Thus, the domain of the function is all positive real numbers \((0, \infty)\).

Step 2: Determine the Domain

From the properties of the logarithmic function \( \ln(x) \), the domain is limited to strictly positive \( x \) values. Therefore, the domain of \( h(x) = 3 \ln(x) - 9 \) is \( (0, \infty) \).

Step 3: Determine the Range

For \( x > 0 \), \( \ln(x) \) ranges from \(-\infty\) to \( \infty\). Multiplying by 3 and subtracting 9 linearly shifts the range but does not alter its extent. Thus, the range of \( h(x) = 3 \ln(x) - 9 \) is also \( (-\infty, \infty) \).

Step 4: Find the x-intercept

To find the \( x \)-intercept, set \( h(x) = 0 \) and solve for \( x \):\[ 3 \ln(x) - 9 = 0 \] \[ 3 \ln(x) = 9 \] \[ \ln(x) = 3 \] Convert to exponential form: \[ x = e^3 \].Thus, the \( x \)-intercept is \( (e^3, 0) \).

Step 5: Find the y-intercept

Set \( x = 0 \) to find the \( y \)-intercept, but since \( \ln(x) \) is undefined at \( x = 0 \), the \( y \)-intercept does not exist (DNE).

Key Concepts

Domain of a Function

The domain of a function is the set of all possible input values (typically represented by \( x \)) that the function can accept without causing any mathematical issues, such as division by zero or taking the logarithm of a negative number. For the function \( h(x) = 3 \ln(x) - 9 \), since \( \ln(x) \) is only defined for positive values of \( x \), the domain of this function is all real numbers greater than zero, expressed as \( (0, \infty) \). This means that for any positive real number you substitute into \( h(x) \), the function will yield a defined, real-number output.
A simple way to remember the domain rules for a logarithmic function is:

• The logarithmic function \( \ln(x) \) is only defined for \( x > 0 \).
• So, avoid using zero or negative numbers in the function's input.

Understanding the domain of a function is crucial because it informs you which values are permissible to use without causing undefined mathematical expressions.

Range of a Function

The range of a function refers to the set of all possible output values the function can produce as \( x \) runs through its domain. Despite the function's specific transformations, logarithmic functions typically can output all real numbers. For the function \( h(x) = 3 \ln(x) - 9 \), the range covers all real numbers, \( (-\infty, \infty) \).
This can be broken down as follows:

• Initially, \( \ln(x) \) on its own can produce any real number from \(-\infty \) to \( \infty \) as x moves from just above 0 to infinity.
• Multiplying by 3 expands or stretches the range but still includes every value in the real number line over time because log's original domain affects how quickly it hits large or small numbers.
• Finally, subtracting 9 shifts every output down, still covering all real values eventually.

This comprehensive range shows that regardless of the input, \( h(x) \) can approach any real number as output.

Intercepts of a Function

Intercepts are where the graph of a function intersects the axes. For this function, we can find both the \( x \)-intercept and the \( y \)-intercept:

• \( x \)-Intercept: The \( x \)-intercept occurs where the function value is zero. Set \( h(x) = 0 \):
  \[ 3 \ln(x) - 9 = 0 \]
  Solve for \( x \):
  \[ 3 \ln(x) = 9 \rightarrow \ln(x) = 3 \rightarrow x = e^3 \]
  Hence, the \( x \)-intercept is \((e^3, 0)\).
• \( y \)-Intercept: The \( y \)-intercept occurs where \( x = 0 \), but in this case, \( \ln(x) \) is undefined at \( x = 0 \). Therefore, there is no \( y \)-intercept (DNE).

Understanding intercepts helps visualize where the graph crosses the axes and gives insights into the function's behavior in both positive and negative domains.