Question

The range of the function $f(x)=\log_{e}(4x^{2}-4x+1)$ where $x \ne \frac{1}{2}$ is ________.

• A. $(-\infty,0)$
• B. $[0,\infty)$
• C. $(0,\infty)$
• D. $(-\infty,0]$
• E. $(-\infty,\infty)$

Hint:

Logarithmic functions typically have a range of all real numbers.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
The range of a function is the set of all possible output values (y-values) it can produce. We need to analyze the argument of the logarithm to determine what values it can take, which in turn determines the range of the logarithmic function.
Step 2: Key Formula or Approach
The standard logarithmic function \(g(z) = \log_{a}(z)\) is defined for \(z>0\). The range of \(g(z)\) is all real numbers, i.e., \((-\infty, \infty)\). Our strategy is to simplify the argument of the given function and find the set of values it can take.
Step 3: Detailed Explanation
1. Simplify the argument of the logarithm.
The given function is \(f(x) = \log_{a}(4x^2 - 4x + 1)\).
The argument is the quadratic expression \(4x^2 - 4x + 1\). We can recognize this as a perfect square trinomial.
\[ 4x^2 - 4x + 1 = (2x)^2 - 2(2x)(1) + (1)^2 = (2x - 1)^2 \] So, the function can be rewritten as:
\[ f(x) = \log_{a}((2x - 1)^2) \] 2. Analyze the values the argument can take.
Let \(z = (2x - 1)^2\).
The problem states that \(x \neq \frac{1}{2}\). This is the natural domain restriction, because if \(x = \frac{1}{2}\), then \(z = (2(\frac{1}{2}) - 1)^2 = (1 - 1)^2 = 0\), and \(\log_{a}(0)\) is undefined.
For any real number \(x\) other than \(\frac{1}{2}\), the term \((2x - 1)\) will be a non-zero real number.
When we square any non-zero real number, the result is always a strictly positive real number.
Therefore, \(z = (2x - 1)^2>0\).
The argument \(z\) can take any value in the interval \((0, \infty)\).
3. Determine the range of the function.
Our function is now effectively \(f(x) = \log_{a}(z)\), where \(z\) can be any positive real number.
The range of the basic logarithmic function \(\log_{a}(z)\) for \(z \in (0, \infty)\) is the set of all real numbers, \(\mathbb{R}\).
Thus, the range of \(f(x)\) is \((-\infty, \infty)\).
Step 4: Final Answer
The range of the function \(f(x)\) is \((-\infty, \infty)\).