Math Problem Statement
range of 1/(ln(x)-1)
Solution
To find the range of the function , we need to consider the domain of the logarithmic function and the behavior of the expression as varies.
Step 1: Identify the domain of the function
The logarithmic function is defined for , and when . Additionally, approaches as approaches , and approaches as increases.
However, for the function to be defined, the denominator must not be zero: Thus, the domain of is except .
Step 2: Analyze the behavior of the function
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As approaches 0 from the right (): approaches , so also approaches . Therefore, approaches 0 from the negative side.
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As increases towards from the left (): approaches 1 from below, making approach 0 from the negative side, so approaches .
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As increases beyond (): becomes positive, and as increases, decreases towards 0 from the positive side.
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At large (): increases without bound, so becomes very large, and approaches 0 from the positive side.
Step 3: Combine the information to determine the range
Considering the behavior of the function, the range of excludes the value 0, but it includes all other real numbers. Specifically, can take any value from to (excluding 0) when is just less than , and any value from (excluding 0) to when is just greater than .
Conclusion:
The range of is .
Would you like a more detailed explanation or have any questions?
Here are some related questions to consider:
- What is the derivative of the function ?
- How would the graph of look, and what are its asymptotes?
- What happens to the function as approaches ?
- Can you find the inverse of ?
- How does the function compare to ?
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
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