Math Problem Statement

Determine the type of each function, sketch the graph, and find its domain & range for the following functions: f(x) = x^2 - 1, f(x) = sqrt(x - 2), f(x) = sqrt(x^2 - 2), f(x) = 1 / (x - 2), f(x) = 1 / sqrt(x), f(x) = sqrt(4 - x^2), f(x) = |x - 1| + 1.

Solution

The image shows a math exercise that involves identifying the type of each function and finding its domain and range. The functions are labeled 1 to 7. Let's break down each function, identify its type, and determine the domain and range:

1. f(x)=x21f(x) = x^2 - 1

  • Type: Quadratic function (since it's a polynomial of degree 2).
  • Domain: All real numbers R\mathbb{R} (since any real number can be squared).
  • Range: The function has a minimum value of 1-1 (at x=0x = 0) and increases to infinity. So, Range=[1,)\text{Range} = [-1, \infty).

2. f(x)=x2f(x) = \sqrt{x - 2}

  • Type: Square root function.
  • Domain: x20x - 2 \geq 0, meaning x2x \geq 2. Thus, Domain=[2,)\text{Domain} = [2, \infty).
  • Range: The square root function produces only non-negative outputs, so the range is [0,)[0, \infty).

3. f(x)=x22f(x) = \sqrt{x^2 - 2}

  • Type: Square root of a quadratic function.
  • Domain: x220x^2 - 2 \geq 0, which simplifies to x2x \leq -\sqrt{2} or x2x \geq \sqrt{2}. So, Domain=(,2][2,)\text{Domain} = (-\infty, -\sqrt{2}] \cup [\sqrt{2}, \infty).
  • Range: The range starts at 0 and increases to infinity. Therefore, Range=[0,)\text{Range} = [0, \infty).

4. f(x)=1x2f(x) = \dfrac{1}{x - 2}

  • Type: Rational function.
  • Domain: The function is undefined when the denominator is zero, i.e., x2x \neq 2. Thus, Domain=(,2)(2,)\text{Domain} = (-\infty, 2) \cup (2, \infty).
  • Range: The function's values can approach both negative and positive infinity but never reach 0, so Range=(,0)(0,)\text{Range} = (-\infty, 0) \cup (0, \infty).

5. f(x)=1xf(x) = \dfrac{1}{\sqrt{x}}

  • Type: Rational function with a square root.
  • Domain: The square root is defined for x0x \geq 0, and the denominator cannot be zero, so x>0x > 0. Thus, Domain=(0,)\text{Domain} = (0, \infty).
  • Range: The function approaches infinity as x0+x \to 0^+ and decreases towards 0 as xx \to \infty. So, Range=(0,)\text{Range} = (0, \infty).

6. f(x)=4x2f(x) = \sqrt{4 - x^2}

  • Type: Square root of a quadratic function (circular segment).
  • Domain: 4x204 - x^2 \geq 0, so 2x2-2 \leq x \leq 2. Thus, Domain=[2,2]\text{Domain} = [-2, 2].
  • Range: The function produces values between 0 and 2, inclusive, so Range=[0,2]\text{Range} = [0, 2].

7. f(x)=x1+1f(x) = |x - 1| + 1

  • Type: Absolute value function.
  • Domain: The absolute value function is defined for all real numbers, so Domain=R\text{Domain} = \mathbb{R}.
  • Range: The minimum value of the function is 1 (when x=1x = 1), and it increases without bound. Thus, Range=[1,)\text{Range} = [1, \infty).

Would you like further clarification on any of these functions?

Additional Questions:

  1. How do you graph a quadratic function like f(x)=x21f(x) = x^2 - 1?
  2. Can you explain how to solve for the domain of functions involving square roots?
  3. What is the relationship between the degree of a polynomial and the shape of its graph?
  4. How does the range of a rational function differ from that of a quadratic function?
  5. What are some key differences between absolute value functions and quadratic functions?

Tip:

When finding the domain of a function, always check for values that make the expression inside a square root negative or the denominator equal to zero, as these create restrictions.

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Math Problem Analysis

Mathematical Concepts

Quadratic functions
Square root functions
Rational functions
Absolute value functions

Formulas

f(x) = x^2 - 1
f(x) = sqrt(x - 2)
f(x) = sqrt(x^2 - 2)
f(x) = 1 / (x - 2)
f(x) = 1 / sqrt(x)
f(x) = sqrt(4 - x^2)
f(x) = |x - 1| + 1

Theorems

Domain restrictions from square roots and rational functions

Suitable Grade Level

Grades 10-12