How do you determine whether a function is odd, even, or neither: #h(x)= -x^3/(3x^2-9)#?

Question

Anna Allen · Accepted Answer

Odd

We substitude #x# with #-x# : #h(-x)=-(-x)^3/(3(-x)^2-9)=-(-x^3)/(3x^2-9)=x^3/(3x^2-9)=# #-h(x)# Now because #h(-x)=-h(x)# the function is odd.

Benjamin Achtenberg · Answer

#h(x)# is an odd function.

We apply the subsequent prerequisite: # { (f(-x)=f(x)), (f(-x)=-f(x)) :} => {: (f " is even"), (f " is odd") :}# Thus, for the specified function: # h(x) = -(x^3)/(3x^2-9) # And so: # h(-x) = -((-x)^3)/(3(-x)^2-9) # # " " = -(-x^3)/(3x^2-9) # # " " = (x^3)/(3x^2-9) # # " " = -h(x) # And we conclude that #h(x)# is an odd function. Odd functions have rotational symmetry about the origin, so we can visually confirm this: graph{-(x^3)/(3x^2-9) [-20, 20, -10, 10]}

Greyson Brown · Answer

undefined To determine whether a function is odd, even, or neither, you evaluate the function $ h(x) $ at $ -x $ and observe if it satisfies the properties of odd or even functions.

1. **Odd Function**: A function $ h(x) $ is odd if $ h(-x) = -h(x) $ for all $ x $ in its domain.

2. **Even Function**: A function $ h(x) $ is even if $ h(-x) = h(x) $ for all $ x $ in its domain.

3. **Neither Odd nor Even**: If neither of the above conditions is met, then the function is neither odd nor even.

For the function $ h(x) = -\frac{x^3}{3x^2 - 9} $:

Evaluate $ h(-x) $:

$$ h(-x) = -\frac{(-x)^3}{3(-x)^2 - 9} $$
$$ h(-x) = -\frac{-x^3}{3x^2 - 9} $$

Compare with $ h(x) $:

$$ -h(x) = -\left(-\frac{x^3}{3x^2 - 9}\right) = \frac{x^3}{3x^2 - 9} $$

Since $ h(-x) = h(x) $, the function $ h(x) $ is an even function.