Math Problem Statement

Find the domain and the range of the following real functions: (a) 𝑓(π‘₯) = (π‘₯2 βˆ’ 4)/(x+2)

(f) 𝑓(π‘₯)=|π‘₯|βˆ’π‘₯ /(2π‘₯)

(g)𝑓(π‘₯)= π‘₯βˆ’3/( 2π‘₯ + 1)

(h) 𝑓(π‘₯) = 1 /(1βˆ’π‘₯2)

(i) 𝑓(π‘₯) = √π‘₯2 βˆ’ 4

(j) 𝑓(π‘₯)= π‘₯+2/( |π‘₯ + 2|)

Solution

Let's analyze each function step by step to find the domain (the set of all possible input values) and the range (the set of all possible output values).


(a) f(x)=x2βˆ’4x+2f(x) = \frac{x^2 - 4}{x + 2}

Domain:

  • The denominator x+2x + 2 must not be zero, so xβ‰ βˆ’2x \neq -2.
  • Thus, the domain is: (βˆ’βˆž,βˆ’2)βˆͺ(βˆ’2,∞)\boxed{(-\infty, -2) \cup (-2, \infty)}

Simplification:

  • The numerator x2βˆ’4x^2 - 4 can be factored as (xβˆ’2)(x+2)(x - 2)(x + 2), so the function simplifies: f(x)=(xβˆ’2)(x+2)x+2f(x) = \frac{(x - 2)(x + 2)}{x + 2} For xβ‰ βˆ’2x \neq -2, this simplifies to: f(x)=xβˆ’2f(x) = x - 2

Range:

  • Since f(x)=xβˆ’2f(x) = x - 2 for all xβ‰ βˆ’2x \neq -2, the range is all real numbers: (βˆ’βˆž,∞)\boxed{(-\infty, \infty)}

(f) f(x)=∣xβˆ£βˆ’x2xf(x) = \frac{|x| - x}{2x}

Domain:

  • The denominator 2x2x must not be zero, so xβ‰ 0x \neq 0.
  • Thus, the domain is: (βˆ’βˆž,0)βˆͺ(0,∞)\boxed{(-\infty, 0) \cup (0, \infty)}

Simplification:

  • For x>0x > 0, ∣x∣=x|x| = x, so f(x)=xβˆ’x2x=0f(x) = \frac{x - x}{2x} = 0.
  • For x<0x < 0, ∣x∣=βˆ’x|x| = -x, so f(x)=βˆ’xβˆ’x2x=βˆ’2x2x=βˆ’1f(x) = \frac{-x - x}{2x} = \frac{-2x}{2x} = -1.

Range:

  • For x>0x > 0, f(x)=0f(x) = 0.
  • For x<0x < 0, f(x)=βˆ’1f(x) = -1.
  • Therefore, the range is: {βˆ’1,0}\boxed{\{-1, 0\}}

(g) f(x)=xβˆ’32x+1f(x) = \frac{x - 3}{2x + 1}

Domain:

  • The denominator 2x+1β‰ 02x + 1 \neq 0, so xβ‰ βˆ’12x \neq -\frac{1}{2}.
  • Thus, the domain is: (βˆ’βˆž,βˆ’12)βˆͺ(βˆ’12,∞)\boxed{(-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty)}

Range:

  • ( f(x

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Math Problem Analysis

Mathematical Concepts

Functions
Domain and Range
Absolute Value
Square Roots
Rational Functions

Formulas

f(x) = (x^2 - 4)/(x + 2)
f(x) = |x| βˆ’ x / (2x)
f(x) = (x - 3)/(2x + 1)
f(x) = 1/(1 - x^2)
f(x) = √(x^2 - 4)
f(x) = (x + 2)/|x + 2|

Theorems

Domain exclusion for denominators
Properties of absolute value
Range determination for rational functions

Suitable Grade Level

Grades 10-12