Math Problem Statement

Let f(x) = (2 - e^(-x) / (e^(-x) - 1))^(1/2). Find (a) the domain of f(x), (b) the inverse of f(x), and (c) the domain of the inverse.

Solution

Let's break down the problem step by step.

We are given the function:

f(x)=(2exex1)1/2f(x) = \left( \frac{2 - e^{-x}}{e^{-x} - 1} \right)^{1/2}

The problem asks us to:

  • (a) Find the domain of f(x)f(x)
  • (b) Find the inverse f1f^{-1}
  • (c) Find the domain of the inverse f1f^{-1}

Part (a): Find the domain of f(x)f(x)

The domain of a function refers to the set of values for which the function is defined. For this function, there are two main considerations:

  1. The fraction inside the square root must be defined, which means the denominator cannot be zero: ex10e^{-x} - 1 \neq 0 Solving this: ex1x0x0e^{-x} \neq 1 \quad \Rightarrow \quad -x \neq 0 \quad \Rightarrow \quad x \neq 0 So, x0x \neq 0.

  2. The expression inside the square root must be non-negative (since we're dealing with real numbers): 2exex10\frac{2 - e^{-x}}{e^{-x} - 1} \geq 0 Let's examine when this fraction is non-negative:

    • The numerator, 2ex2 - e^{-x}, is positive when ex<2e^{-x} < 2, which is true for all xx since exe^{-x} is always less than 2.
    • The denominator, ex1e^{-x} - 1, is positive when ex>1e^{-x} > 1, i.e., x<0x < 0, and negative when ex<1e^{-x} < 1, i.e., x>0x > 0.

    Thus, the fraction is non-negative for x<0x < 0 and negative for x>0x > 0.

Conclusion: The domain of f(x)f(x) is x<0x < 0, or (,0)(-\infty, 0).

Part (b): Find the inverse f1(x)f^{-1}(x)

To find the inverse, we first rewrite the equation for f(x)f(x):

y=(2exex1)1/2y = \left( \frac{2 - e^{-x}}{e^{-x} - 1} \right)^{1/2}

Now square both sides to get rid of the square root:

y2=2exex1y^2 = \frac{2 - e^{-x}}{e^{-x} - 1}

Now, solve for xx:

  1. Multiply both sides by ex1e^{-x} - 1: y2(ex1)=2exy^2(e^{-x} - 1) = 2 - e^{-x}
  2. Distribute y2y^2 on the left-hand side: y2exy2=2exy^2 e^{-x} - y^2 = 2 - e^{-x}
  3. Bring all terms involving exe^{-x} to one side: y2ex+ex=2+y2y^2 e^{-x} + e^{-x} = 2 + y^2
  4. Factor out exe^{-x} on the left-hand side: ex(y2+1)=2+y2e^{-x}(y^2 + 1) = 2 + y^2
  5. Solve for exe^{-x}: ex=2+y2y2+1e^{-x} = \frac{2 + y^2}{y^2 + 1}
  6. Take the natural logarithm of both sides: x=ln(2+y2y2+1)-x = \ln \left( \frac{2 + y^2}{y^2 + 1} \right)
  7. Solve for xx: x=ln(2+y2y2+1)x = -\ln \left( \frac{2 + y^2}{y^2 + 1} \right)

Thus, the inverse function is:

f1(y)=ln(2+y2y2+1)f^{-1}(y) = -\ln \left( \frac{2 + y^2}{y^2 + 1} \right)

Part (c): Find the domain of the inverse f1(x)f^{-1}(x)

The domain of the inverse is the range of the original function. From the form of f(x)f(x), we see that:

  • The square root ensures that f(x)0f(x) \geq 0.
  • As xx \to -\infty, f(x)0f(x) \to 0.
  • As x0x \to 0^-, f(x)f(x) \to \infty.

Thus, the range of f(x)f(x) is [0,)[0, \infty), meaning the domain of the inverse function f1(x)f^{-1}(x) is [0,)[0, \infty).


Would you like more detailed steps or further clarification? Here are some related questions for deeper understanding:

  1. How do you identify when a fraction is non-negative?
  2. What does the range of a function tell you about its inverse?
  3. How can we find the domain of more complex expressions like f(x)f(x)?
  4. Why do we take the logarithm when solving for the inverse?
  5. Can the function f(x)f(x) ever be negative?

Tip: Always check for restrictions on the function (like square roots or denominators) when determining the domain.

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Math Problem Analysis

Mathematical Concepts

Functions
Domain and Range
Inverse Functions
Logarithms

Formulas

f(x) = (2 - e^(-x)) / (e^(-x) - 1)
Inverse: f^(-1)(y) = -ln((2 + y^2) / (y^2 + 1))

Theorems

Square root and non-negative constraints
Natural logarithm properties for inverse functions

Suitable Grade Level

Undergraduate (Calculus/Pre-Calculus)