Math Problem Statement

Determine whether each function is even, odd, or neither for the following functions: f(x) = 1/(4x^3), g(x) = 3x |x|, h(x) = √(8x^2 + 1).

Solution

The image contains three functions, f(x)f(x), g(x)g(x), and h(x)h(x), and asks to determine if each function is even, odd, or neither.

  1. For f(x)=14x3f(x) = \frac{1}{4x^3}:

    • A function is odd if f(x)=f(x)f(-x) = -f(x).
    • Let's test f(x)f(-x): f(x)=14(x)3=14x3=14x3=f(x)f(-x) = \frac{1}{4(-x)^3} = \frac{1}{-4x^3} = -\frac{1}{4x^3} = -f(x)
    • Since f(x)=f(x)f(-x) = -f(x), the function f(x)f(x) is odd.
  2. For g(x)=3xxg(x) = 3x |x|:

    • A function is odd if g(x)=g(x)g(-x) = -g(x).
    • Let's test g(x)g(-x): g(x)=3(x)x=3(x)(x)=3xx=g(x)g(-x) = 3(-x) |-x| = 3(-x)(x) = -3x |x| = -g(x)
    • Since g(x)=g(x)g(-x) = -g(x), the function g(x)g(x) is also odd.
  3. For h(x)=8x2+1h(x) = \sqrt{8x^2 + 1}:

    • A function is even if h(x)=h(x)h(-x) = h(x).
    • Let's test h(x)h(-x): h(x)=8(x)2+1=8x2+1=h(x)h(-x) = \sqrt{8(-x)^2 + 1} = \sqrt{8x^2 + 1} = h(x)
    • Since h(x)=h(x)h(-x) = h(x), the function h(x)h(x) is even.

Summary:

  • f(x)f(x) is odd.
  • g(x)g(x) is odd.
  • h(x)h(x) is even.

Would you like further details or have any questions?

Here are 5 related questions:

  1. How can we determine if a function is neither even nor odd?
  2. What happens when a function is both even and odd?
  3. Can a polynomial function of degree 3 be even?
  4. How does the presence of absolute values affect the symmetry of a function?
  5. What other methods can be used to verify if a function is even or odd?

Tip: For any rational or radical function, always check the behavior of the function under input sign changes to determine its symmetry properties.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Even and Odd Functions
Function Symmetry
Algebraic Functions

Formulas

f(x) = 1/(4x^3)
g(x) = 3x |x|
h(x) = √(8x^2 + 1)

Theorems

Properties of Even and Odd Functions

Suitable Grade Level

Grades 10-12