Problem 21 - College Algebra

Question

For the following exercises, state the domain, range, and \(x\) -and \(y\) -intercepts, if they do not exist, write DNE. $$h(x)=\log _{4}(x-1)+1$$

Step-by-Step Solution

Answer

Domain: (1, ∞); Range: (-∞, ∞); x-intercept: x = 5/4; y-intercept: DNE.

Step 1: Find the Domain

The function is defined as a logarithm, \( h(x) = \log_4(x-1) + 1 \). The argument of the logarithm, \( (x-1) \), must be greater than 0. Set up the inequality: \( x-1 > 0 \). Solving this gives \( x > 1 \). Therefore, the domain of the function is \( (1, \infty) \).

Step 2: Determine the Range

A logarithmic function can take any real value based on its argument. Therefore, \( h(x) = \log_4(x-1) + 1 \) can take all real numbers. Hence, the range is \( (-\infty, \infty) \).

Step 3: Find the x-intercept

To find the \( x \)-intercept, set \( h(x) = 0 \). So, \( \log_4(x-1) + 1 = 0 \). Subtract 1 from both sides: \( \log_4(x-1) = -1 \). Convert this logarithmic equation to its exponential form: \( x-1 = 4^{-1} = \frac{1}{4} \). Therefore, \( x = 1 + \frac{1}{4} = \frac{5}{4} \). The \( x \)-intercept is at \( x = \frac{5}{4} \).

Step 4: Find the y-intercept

Substitute \( x = 0 \) into the function to find the \( y \)-intercept: \( h(0) = \log_4(0-1) + 1 \). Since the logarithm of a negative number is undefined, there is no \( y \)-intercept for this function. Thus, write \( \text{DNE} \) for the \( y \)-intercept.

Key Concepts

Domain and Range of Logarithmic Functions

When working with logarithmic functions, it's crucial to identify their domain and range. Let's consider the function given in the exercise, \( h(x) = \log_4(x-1) + 1 \). Understanding the domain means identifying all possible input values \( x \) for which the function is defined. The crucial part of a logarithmic function is its argument, since it must always be positive. Hence, for the function \( h(x) \), the argument \( x-1 \) must be greater than zero.

• To solve \( x-1 > 0 \), add 1 to both sides, giving you \( x > 1 \).

This tells us that the domain of the function is \( (1, \infty) \). This means that \( x \) can be any real number greater than 1. For the range, logarithmic functions can produce any real number. This particular function with the added constant of 1 means the range remains unchanged, spanning all real numbers: \( (-\infty, \infty) \).

X-intercepts of Logarithmic Functions

Finding the \(x\)-intercepts involves determining where the graph of the function crosses the \(x\)-axis. For any function, this is where the function's value equals zero. For our logarithmic function \( h(x) = \log_4(x-1) + 1 \), set \( h(x) = 0 \) and solve for \( x \).

• Start by subtracting 1 on both sides: \( \log_4(x-1) = -1 \).
• Convert the logarithmic equation to its equivalent exponential form: \( x-1 = 4^{-1} = \frac{1}{4} \).
• Add 1 to \( \frac{1}{4} \) to find \( x = \frac{5}{4} \).

Thus, the \(x\)-intercept is at \( x = \frac{5}{4} \), indicating that the graph meets the \(x\)-axis at this point. This is where the function value becomes zero. Remember, finding intercepts can be critical for understanding the behavior of the graph.

Y-intercepts of Logarithmic Functions

The \(y\)-intercepts occur where the graph crosses the \(y\)-axis, meaning when \(x = 0\). However, in logarithmic functions like \( h(x) = \log_4(x - 1) + 1 \), if substituting \( x = 0 \) into the function results in an undefined expression, then no \(y\)-intercept exists.

• Substituting \( x = 0 \) results in \( h(0) = \log_4(0-1) + 1 \).
• The term \( \log_4(-1) \) is undefined because a logarithm cannot take a negative argument.

Hence, for this function, substituting \( x = 0 \) gives a value that doesn't exist within the real number domain, meaning the \(y\)-intercept does not exist, or DNE. It's important to always check if the arguments meet the conditions for definition to determine any potential \(y\)-intercepts effectively.