Math Problem Statement

f(x)=\sqrt((x^(2)-x)/(x-1))

Solution

The function you provided is:

f(x)=x2xx1f(x) = \sqrt{\frac{x^2 - x}{x - 1}}

Let's analyze this function and determine its behavior in terms of domain and any simplifications.


1. Simplifying the expression under the square root:

First, simplify the fraction x2xx1\frac{x^2 - x}{x - 1}. Notice that the numerator x2xx^2 - x can be factored:

x2x=x(x1)x^2 - x = x(x - 1)

Thus:

x2xx1=x(x1)x1.\frac{x^2 - x}{x - 1} = \frac{x(x - 1)}{x - 1}.

For x1x \neq 1, we can cancel the common factor (x1)(x - 1), leaving:

x2xx1=x(for x1).\frac{x^2 - x}{x - 1} = x \quad (\text{for } x \neq 1).

Therefore, the function becomes:

f(x)=x,for x1.f(x) = \sqrt{x}, \quad \text{for } x \neq 1.


2. Domain analysis:

Since we are dealing with a square root function, the expression inside the square root (which simplifies to xx) must be non-negative:

x0.x \geq 0.

Additionally, we must consider that the original fraction x2xx1\frac{x^2 - x}{x - 1} is undefined at x=1x = 1. Therefore, the final domain is:

x0andx1.x \geq 0 \quad \text{and} \quad x \neq 1.

Thus, the domain of f(x)f(x) is:

Domain: [0,1)(1,).\text{Domain: } [0, 1) \cup (1, \infty).


3. Final expression of the function:

With the simplification, the function can be rewritten as:

f(x)=x,for x1.f(x) = \sqrt{x}, \quad \text{for } x \neq 1.


4. Key Points about f(x)f(x):

  • At x=0x = 0: f(0)=0=0f(0) = \sqrt{0} = 0.
  • As xx \to \infty: f(x)f(x) \to \infty.
  • At x=1x = 1: The function is undefined because of the original expression's denominator being zero.

5. Summary:

The simplified function is f(x)=xf(x) = \sqrt{x}, with the domain x0x \geq 0 but excluding x=1x = 1.


Would you like further clarification or details about graphing this function, continuity, or related properties?


Here are five follow-up questions:

  1. How do we find the range of the function f(x)=xf(x) = \sqrt{x}?
  2. What happens if the function had a cube root instead of a square root?
  3. How would the graph of f(x)=x2xx1f(x) = \sqrt{\frac{x^2 - x}{x - 1}} look?
  4. Can we determine if the function is continuous on its domain?
  5. How would adding a constant inside or outside the square root affect the domain?

Tip: Always double-check for restrictions when dealing with rational functions and square roots to ensure proper domain analysis.

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Math Problem Analysis

Mathematical Concepts

Algebra
Function Simplification
Square Roots
Domain of Functions

Formulas

f(x) = \sqrt{\frac{x^2 - x}{x - 1}}
Domain restriction for square root: expression under root >= 0

Theorems

Properties of Rational Functions
Square Root Function Domain Restrictions

Suitable Grade Level

Grades 11-12