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Question
Math
Posted 7 months ago

Let ff be the function given by f(x)=ln⁡xx−1\boldsymbol{f}(\boldsymbol{x})=\ln \frac{\boldsymbol{x}}{\boldsymbol{x}-1}.
(a) What is the domain of ff ?
(b) Find the value of the derivative of ff at x=−1x=-1.
(c) Write an expression for f−1(x)f^{-1}(x), where f−1f^{-1} denotes the inverse function of ff.
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Answer from Sia
Posted 7 months ago
Solution by Steps
step 1
To find the domain of the function f(x)=ln⁡(xx−1)f(x) = \ln \left( \frac{x}{x-1} \right), we need to determine where the argument of the logarithm is positive
step 2
The argument xx−1\frac{x}{x-1} is positive when x>1x > 1 or x<0x < 0
step 3
Therefore, the domain of ff is (−∞,0)∪(1,∞)(-\infty, 0) \cup (1, \infty)
Answer
The domain of ff is (−∞,0)∪(1,∞)(-\infty, 0) \cup (1, \infty).
Key Concept
Domain of a logarithmic function
Explanation
The domain of a logarithmic function is determined by the positivity of its argument.


 Part (b)
step 1
To find the derivative of f(x)=ln⁡(xx−1)f(x) = \ln \left( \frac{x}{x-1} \right) at x=−1x = -1, we use the chain rule
step 2
The derivative of ln⁡(xx−1)\ln \left( \frac{x}{x-1} \right) is ddx(ln⁡(xx−1))=1xx−1⋅ddx(xx−1)\frac{d}{dx} \left( \ln \left( \frac{x}{x-1} \right) \right) = \frac{1}{\frac{x}{x-1}} \cdot \frac{d}{dx} \left( \frac{x}{x-1} \right)
step 3
Simplifying, we get ddx(xx−1)=(x−1)⋅1−x⋅1(x−1)2=−1(x−1)2\frac{d}{dx} \left( \frac{x}{x-1} \right) = \frac{(x-1) \cdot 1 - x \cdot 1}{(x-1)^2} = \frac{-1}{(x-1)^2}
step 4
Therefore, the derivative is 1xx−1⋅−1(x−1)2=−1x(x−1)\frac{1}{\frac{x}{x-1}} \cdot \frac{-1}{(x-1)^2} = \frac{-1}{x(x-1)}
step 5
Evaluating at x=−1x = -1, we get −1−1(−1−1)=−12=−12\frac{-1}{-1(-1-1)} = \frac{-1}{2} = -\frac{1}{2}
Answer
The value of the derivative of ff at x=−1x = -1 is −12-\frac{1}{2}.
Key Concept
Derivative of a logarithmic function
Explanation
The derivative of a logarithmic function can be found using the chain rule and simplifying the resulting expression.


 Part (c)
step 1
To find the inverse function of f(x)=ln⁡(xx−1)f(x) = \ln \left( \frac{x}{x-1} \right), we start by setting y=ln⁡(xx−1)y = \ln \left( \frac{x}{x-1} \right)
step 2
Exponentiating both sides, we get ey=xx−1e^y = \frac{x}{x-1}
step 3
Solving for xx, we get x=eyey−1x = \frac{e^y}{e^y - 1}
Answer
The inverse function of ff is f−1(x)=exex−1f^{-1}(x) = \frac{e^x}{e^x - 1}.
Key Concept
Inverse of a logarithmic function
Explanation
The inverse of a logarithmic function can be found by exponentiating and solving for the original variable.

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