How do you determine if #f(x)= (-x³)/(3x²-4)# is an even or odd function?

Question

Serenity Johnson · Accepted Answer

See below

OVERALL, TALKING _ There are assessments available: Function EVEN: f(-x) = f(x) so, for example, the ubiquitous cosine function is even as #cos (-x) = cos x#. Generally speaking, even functions reflect across the y-axis. Function of ODD: f(-x) = - f(x) the equally ubiquitous sine function is odd as #sin(- x) = - sin x#. Odd functions show symmetry in terms of a 180 deg rotation about the origin. But they don't have to pass through the origin, eg #f(x) = 1/x, f(-x) = - 1/x = - f(x) # is odd. NB a function can be neither odd or even. The #ln x# and #e^x# functions, for example. THE QUESTION IN ACT so here we test #f(x)= (-x³)/(3x²-4)# by looking at #f(-x)# which is #f(-x) = (-(-x)³)/(3(-x)²-4)# #= (x³)/(3x²-4) = - f(x)# This is peculiar. _ A PICTURE _ graph{ (-x^3)/(3x^2-4) [-18.02, 18.02, -9.01, 9.01]}

Ariana Alvarez · Answer

undefined To determine if a function $ f(x) $ is even or odd, we need to check if it satisfies the following properties:

1. Even Function: $ f(x) = f(-x) $ for all $ x $ in the function's domain.
2. Odd Function: $ f(x) = -f(-x) $ for all $ x $ in the function's domain.

For the function $ f(x) = \frac{-x^3}{3x^2 - 4} $, we'll evaluate both $ f(x) $ and $ f(-x) $.

If $ f(x) = f(-x) $, then the function is even. If $ f(x) = -f(-x) $, then the function is odd.

Let's evaluate $ f(x) $ and $ f(-x) $:

$$ f(x) = \frac{-x^3}{3x^2 - 4} $$

$$ f(-x) = \frac{-(-x)^3}{3(-x)^2 - 4} = \frac{-(-x)^3}{3x^2 - 4} = \frac{-(-x^3)}{3x^2 - 4} = \frac{x^3}{3x^2 - 4} $$

Now, we'll compare $ f(x) $ and $ f(-x) $:

$$ f(x) \neq f(-x) $$

$$ f(x) \neq -f(-x) $$

Since neither $ f(x) = f(-x) $ nor $ f(x) = -f(-x) $ holds true for all $ x $ in the function's domain, the function $ f(x) = \frac{-x^3}{3x^2 - 4} $ is neither even nor odd.