Math Problem Statement

range of 1/(ln(x)-1)

Solution

To find the range of the function f(x)=1ln(x)1f(x) = \frac{1}{\ln(x) - 1}, we need to consider the domain of the logarithmic function and the behavior of the expression as ln(x)\ln(x) varies.

Step 1: Identify the domain of the function

The logarithmic function ln(x)\ln(x) is defined for x>0x > 0, and ln(x)=0\ln(x) = 0 when x=1x = 1. Additionally, ln(x)\ln(x) approaches -\infty as xx approaches 0+0^+, and ln(x)\ln(x) approaches ++\infty as xx increases.

However, for the function f(x)f(x) to be defined, the denominator ln(x)1\ln(x) - 1 must not be zero: ln(x)10ln(x)1xe\ln(x) - 1 \neq 0 \Rightarrow \ln(x) \neq 1 \Rightarrow x \neq e Thus, the domain of f(x)f(x) is x>0x > 0 except x=ex = e.

Step 2: Analyze the behavior of the function

  • As xx approaches 0 from the right (x0+x \to 0^+): ln(x)\ln(x) approaches -\infty, so ln(x)1\ln(x) - 1 also approaches -\infty. Therefore, f(x)=1ln(x)1f(x) = \frac{1}{\ln(x) - 1} approaches 0 from the negative side.

  • As xx increases towards ee from the left (xex \to e^-): ln(x)\ln(x) approaches 1 from below, making ln(x)1\ln(x) - 1 approach 0 from the negative side, so f(x)f(x) approaches -\infty.

  • As xx increases beyond ee (x>ex > e): ln(x)1\ln(x) - 1 becomes positive, and as xx increases, f(x)f(x) decreases towards 0 from the positive side.

  • At large xx (xx \to \infty): ln(x)\ln(x) increases without bound, so ln(x)1\ln(x) - 1 becomes very large, and f(x)f(x) approaches 0 from the positive side.

Step 3: Combine the information to determine the range

Considering the behavior of the function, the range of f(x)f(x) excludes the value 0, but it includes all other real numbers. Specifically, f(x)f(x) can take any value from -\infty to 00 (excluding 0) when xx is just less than ee, and any value from 00 (excluding 0) to ++\infty when xx is just greater than ee.

Conclusion:

The range of f(x)=1ln(x)1f(x) = \frac{1}{\ln(x) - 1} is (,0)(0,+)(-\infty, 0) \cup (0, +\infty).

Would you like a more detailed explanation or have any questions?

Here are some related questions to consider:

  1. What is the derivative of the function f(x)=1ln(x)1f(x) = \frac{1}{\ln(x) - 1}?
  2. How would the graph of f(x)=1ln(x)1f(x) = \frac{1}{\ln(x) - 1} look, and what are its asymptotes?
  3. What happens to the function f(x)f(x) as xx approaches e2e^2?
  4. Can you find the inverse of f(x)=1ln(x)1f(x) = \frac{1}{\ln(x) - 1}?
  5. How does the function g(x)=1ln(x)+1g(x) = \frac{1}{\ln(x) + 1} compare to f(x)f(x)?

Tip: When analyzing the range of a function, it's often helpful to consider the behavior of the function as the input approaches critical points, such as where the function is undefined or where the function's derivative changes sign.

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Math Problem Analysis

Mathematical Concepts

Calculus
Logarithmic Functions
Function Range

Formulas

f(x) = 1 / (ln(x) - 1)

Theorems

Behavior of Logarithmic Functions

Suitable Grade Level

Grades 11-12