Question

three functions are given below. for each, decide whether it is an even function, an odd function, or neither.
$f(x) = \frac{1}{9x^3}$
\bigcirc even \bigcirc odd \bigcirc neither
$g(x) = \sqrt3{4x^2}$
\bigcirc even \bigcirc odd \bigcirc neither
$h(x) = x|x + 3|$
\bigcirc even \bigcirc odd \bigcirc neither

Explanation:

Step1: Recall the definitions

A function \( f(x) \) is even if \( f(-x)=f(x) \) for all \( x \) in the domain. A function \( f(x) \) is odd if \( f(-x)= - f(x) \) for all \( x \) in the domain.

Step2: Analyze \( f(x)=\frac{1}{9x^{3}} \)

First, find \( f(-x) \):
\( f(-x)=\frac{1}{9(-x)^{3}}=\frac{1}{9(-x^{3})}=-\frac{1}{9x^{3}}=-f(x) \)
So \( f(x) \) is odd.

Step3: Analyze \( g(x)=\sqrt[3]{4x^{2}} \)

Find \( g(-x) \):
\( g(-x)=\sqrt[3]{4(-x)^{2}}=\sqrt[3]{4x^{2}} = g(x) \)
So \( g(x) \) is even.

Step4: Analyze \( h(x)=x|x + 3| \)

Find \( h(-x) \):
\( h(-x)=-x|-x + 3|=-x|3 - x| \)
Now, check if \( h(-x)=h(x) \) or \( h(-x)=-h(x) \)
\( -h(x)=-x|x + 3| \)
Since \( |3 - x|
eq|x + 3| \) in general (e.g., when \( x = 1 \), \( |3-1| = 2 \), \( |1 + 3|=4 \)), \( h(-x)
eq h(x) \) and \( h(-x)
eq -h(x) \)
So \( h(x) \) is neither.

Answer:

• For \( f(x)=\frac{1}{9x^{3}} \): Odd
• For \( g(x)=\sqrt[3]{4x^{2}} \): Even
• For \( h(x)=x|x + 3| \): Neither