Math Problem Statement

Range of the function f(x)=tan-1(√[x]+[-x])+√2-|x| +1/x² is

Solution

We are tasked with finding the range of the function:

f(x)=tan1(x+[x])+2x+1x2f(x) = \tan^{-1}(\sqrt{x} + [-x]) + \sqrt{2} - |x| + \frac{1}{x^2}

Let's break it down step by step:

  1. Inverse tangent (arctan): tan1(y)\tan^{-1}(y) has a range of (π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}) for all real values of yy.

  2. Analysis of x+[x]\sqrt{x} + [-x]: The square root function, x\sqrt{x}, is defined for x0x \geq 0. Therefore, this function is undefined for negative values of xx, limiting the domain of f(x)f(x) to x0x \geq 0.

  3. Behavior of x+[x]\sqrt{x} + [-x] for non-negative xx:

    • For x0x \geq 0, [x]=0[-x] = 0 (since the greatest integer less than or equal to 00 is 0). So the expression simplifies to just tan1(x)\tan^{-1}(\sqrt{x}).

    The range of tan1(x)\tan^{-1}(\sqrt{x}) as xx varies from 00 to \infty is [0,π2)[0, \frac{\pi}{2}).

  4. Additional terms:

    • 2\sqrt{2} is constant and equals approximately 1.414.
    • x|x| for x0x \geq 0 is just xx, so the term x-|x| becomes x-x.
    • The term 1x2\frac{1}{x^2} is always positive for x>0x > 0 and tends to 0 as xx \to \infty, but becomes very large as x0+x \to 0^+.

Considerations for the range:

  • The term tan1(x)\tan^{-1}(\sqrt{x}) is bounded by [0,π2)[0, \frac{\pi}{2}).
  • 2\sqrt{2} shifts the overall function upward by 1.414.
  • The term x-x decreases as xx increases.
  • 1x2\frac{1}{x^2} increases rapidly as xx approaches 0, adding a very large positive value near x=0x = 0.

Behavior at the boundaries:

  • As x0+x \to 0^+:

    • tan1(x)0\tan^{-1}(\sqrt{x}) \to 0,
    • 2=1.414\sqrt{2} = 1.414,
    • x0-|x| \to 0,
    • 1x2\frac{1}{x^2} \to \infty.

    Therefore, as xx approaches 0, f(x)f(x) \to \infty.

  • As xx \to \infty:

    • tan1(x)π2\tan^{-1}(\sqrt{x}) \to \frac{\pi}{2},
    • 2=1.414\sqrt{2} = 1.414,
    • x-|x| \to -\infty,
    • 1x20\frac{1}{x^2} \to 0.

    So, as xx grows large, f(x)f(x) \to -\infty.

Conclusion:

The function f(x)f(x) tends to \infty as x0+x \to 0^+ and -\infty as xx \to \infty. Hence, the range of the function is:

(,)\boxed{(-\infty, \infty)}

Do you need further details or explanations on any step?

Here are five related questions:

  1. What is the range of f(x)=tan1(x)+sin(x)f(x) = \tan^{-1}(x) + \sin(x)?
  2. How do you find the domain of a function involving square roots?
  3. What is the range of the function g(x)=1x2g(x) = \frac{1}{x^2}?
  4. What properties does the function tan1(x)\tan^{-1}(x) have?
  5. What are the critical points of f(x)=x+1x2f(x) = -|x| + \frac{1}{x^2}?

Tip: When working with inverse trigonometric functions, it’s useful to remember their principal value ranges!

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Math Problem Analysis

Mathematical Concepts

Inverse Trigonometric Functions
Square Roots
Absolute Value
Asymptotic Behavior

Formulas

f(x) = tan^(-1)(√x + [-x]) + √2 - |x| + 1/x²
Range of tan^(-1)(y): (-π/2, π/2)
Square root function: √x
Absolute value function: |x|

Theorems

Range of inverse tangent function
Behavior of rational functions as x → 0 and x → ∞

Suitable Grade Level

Grade 11-12 and beyond (advanced high school or early college level)