Question

determine whether each function is even, odd, or neither. f(x)=√(x²−9) g(x)=|x−3| f(x)=x/(x²−1) g(x)=x+x² even odd neither

Explanation:

For \( f(x) = \sqrt{x^2 - 9} \)

Step1: Recall the definitions

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

Step2: Compute \( f(-x) \)

Substitute \( -x \) into \( f(x) \):
\( f(-x)=\sqrt{(-x)^2 - 9} \)
Since \( (-x)^2=x^2 \), this simplifies to \( f(-x)=\sqrt{x^2 - 9} \)

Step3: Compare with \( f(x) \)

We see that \( f(-x)=\sqrt{x^2 - 9}=f(x) \).

For \( g(x) = |x - 3| \)

Step1: Compute \( g(-x) \)

Substitute \( -x \) into \( g(x) \):
\( g(-x)=|-x - 3|=|-(x + 3)|=|x + 3| \)

Step2: Compare with \( g(x) \) and \( -g(x) \)

\( g(x)=|x - 3| \), and \( -g(x)=-|x - 3| \).
\( |x + 3|
eq|x - 3| \) (e.g., if \( x = 0 \), \( |0 + 3| = 3 \), \( |0 - 3| = 3 \)? Wait, \( x = 0 \): \( g(-0)=g(0)=|0 - 3| = 3 \), \( |-0 - 3|=|-3| = 3 \)? Wait, no, \( g(-x)=|-x - 3|=| - (x + 3)|=|x + 3| \). For \( x = 1 \): \( g(1)=|1 - 3| = 2 \), \( g(-1)=|-1 - 3|=|-4| = 4 \). \( 4
eq2 \) and \( 4
eq - 2 \). So \( g(-x)
eq g(x) \) and \( g(-x)
eq - g(x) \).

For \( f(x)=\frac{x}{x^2 - 1} \)

Step1: Compute \( f(-x) \)

Substitute \( -x \) into \( f(x) \):
\( f(-x)=\frac{-x}{(-x)^2 - 1}=\frac{-x}{x^2 - 1} \)

Step2: Compare with \( -f(x) \)

\( -f(x)=-\frac{x}{x^2 - 1}=\frac{-x}{x^2 - 1} \)
So \( f(-x)=-f(x) \).

For \( g(x)=x + x^2 \)

Answer:

• \( f(x)=\sqrt{x^2 - 9} \): even
• \( g(x)=|x - 3| \): neither
• \( f(x)=\frac{x}{x^2 - 1} \): odd
• \( g(x)=x + x^2 \): neither