Problem 21 - College Algebra with Corequisite Support

Question

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

Step-by-Step Solution

Answer

Domain: (1, ∞), Range: (-∞, ∞), x-intercept: (5/4, 0), y-intercept: DNE.

Step 1: Determine the Domain

The function given is \(h(x) = \log_{4}(x-1) + 1\). The domain of a logarithmic function \(\log_{b}(x)\) is \(x > 0\). Hence, for our function \(\log_{4}(x-1)\), the argument \((x-1)\) must be greater than 0.Solve for x: \(x - 1 > 0\)\(x > 1\)Thus, the domain is \((1, \infty)\).

Step 2: Determine the Range

The range of a logarithmic function \(\log_{b}(x)\) is all real numbers. Since adding 1 to the logarithm in \(h(x) = \log_{4}(x-1) + 1\) is a vertical shift, the range remains all real numbers, \(( -\infty, \infty)\).

Step 3: Find the x-intercept

The x-intercept occurs when \(h(x) = 0\). So we solve for \(x\):\(0 = \log_{4}(x-1) + 1\)Subtract 1 from both sides:\(-1 = \log_{4}(x-1)\)Convert to exponential form:\(x-1 = 4^{-1} = \frac{1}{4}\)\(x = \frac{1}{4} + 1 = \frac{5}{4}\)The x-intercept is \(\left(\frac{5}{4}, 0\right)\).

Step 4: Find the y-intercept

The y-intercept occurs when \(x = 0\). Plug \(x = 0\) into the equation for \(h(x)\):\(h(0) = \log_{4}(0-1) + 1\)\(h(0) = \log_{4}(-1) + 1\)Since \(\log_{4}(-1)\) is undefined (as the logarithm is only defined for positive arguments), the y-intercept does not exist. Therefore, it is DNE (does not exist).

Key Concepts

Domain of a Function

Understanding the domain of a function is crucial as it tells us all the possible input values (usually represented by 'x') that a function will accept. In the case of logarithmic functions, such as \(h(x) = \log_{4}(x-1) + 1\), the domain is determined by the condition that the argument of the logarithm must be greater than zero. This is because logarithms of non-positive numbers are not defined.

To find the domain for our function, we set \((x-1) > 0\). Solving this inequality gives us \(x > 1\), indicating that \(x\) must be greater than 1. Therefore, the domain of the function is \((1, \infty)\). This means you can plug in any number greater than 1 into the function and it will work perfectly fine, but values of 1 or less won't work.

Range of a Function

The range of a function refers to all the possible outputs \( (y-values) \) that a function can produce. For logarithmic functions, the range is quite broad. Specifically, for \(h(x) = \log_{4}(x-1) + 1\), the range is all real numbers.

Why all real numbers, you ask? That's because logarithmic functions have outputs that can stretch from negative infinity up to positive infinity. In our function, adding 1 to \( \log_{4}(x-1) \) does not change the stretch of the values but shifts the graph vertically by one unit. Therefore, the range remains unchanged, from \(-\infty\) to \(\infty\). This characteristic allows the function to model a variety of real-world situations where outcomes are expected to vary widely.

X-intercept

Finding the x-intercept of a logarithmic function can help us understand where the graph crosses the x-axis, i.e., where the output (y-value) is zero. For \(h(x) = \log_{4}(x-1) + 1\), we find the x-intercept by setting \(h(x) = 0\). This gives us the equation:

• \(0 = \log_{4}(x-1) + 1\)
• Subtract 1 from both sides to isolate the logarithm: \(-1 = \log_{4}(x-1)\)
• Rewrite in exponential form: \(x-1 = 4^{-1}\)
• Solve for \(x\): \(x = \frac{1}{4} + 1 = \frac{5}{4}\)

Thus, the x-intercept is \(\left(\frac{5}{4}, 0\right)\). This means that when \(x\) is \(\frac{5}{4}\), the function outputs \(0\), showing a clear crossing point on the x-axis.

Y-intercept

A y-intercept is the point where the graph of a function crosses the y-axis. It occurs when \(x = 0\). To find the y-intercept of \(h(x) = \log_{4}(x-1) + 1\), we substitute \(x = 0\) into the equation:

• \(h(0) = \log_{4}(0-1) + 1\)

Here, we encounter a problem. The expression \(\log_{4}(-1)\) is undefined as you cannot take the logarithm of a negative number in the real number system. This means the concept of a y-intercept in this function simply does not apply. Consequently, the y-intercept does not exist for this specific logarithmic function. This situation illustrates why understanding domain constraints is essential when exploring function values.