The problem consists of two parts. Part 1: Analyze the parity of two functions. The first function is $g(x) = |x| - \frac{1}{x^2}$. The second function is $h(x) = x\sqrt{x^2 - 1}$. Part 2: Given the function $h(x) = \frac{2x+5}{x+1}$, (a) calculate the image of 3 and -2 by the function $h$, i.e., calculate $h(3)$ and $h(-2)$. (b) determine the antecedent(s) of 5 by the function $h$, i.e., find $x$ such that $h(x) = 5$.

1. Problem Description

The problem consists of two parts.

Part 1: Analyze the parity of two functions.

The first function is $g(x) = |x| - \frac{1}{x^2}$.

The second function is $h(x) = x\sqrt{x^2 - 1}$.

Part 2: Given the function $h(x) = \frac{2x+5}{x+1}$,

(a) calculate the image of 3 and -2 by the function $h$, i.e., calculate $h(3)$ and $h(-2)$.

(b) determine the antecedent(s) of 5 by the function $h$, i.e., find $x$ such that $h(x) = 5$.

2. Solution Steps

Part 1: Parity of functions.

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

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

For $g(x) = |x| - \frac{1}{x^2}$:

$g(-x) = |-x| - \frac{1}{(-x)^2} = |x| - \frac{1}{x^2} = g(x)$.

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

For $h(x) = x\sqrt{x^2 - 1}$:

$h(-x) = (-x)\sqrt{(-x)^2 - 1} = -x\sqrt{x^2 - 1} = -h(x)$.

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

Part 2: Function evaluation and finding antecedents.

(a) To calculate $h(3)$, we substitute $x=3$ into the function $h(x) = \frac{2x+5}{x+1}$:

$h(3) = \frac{2(3)+5}{3+1} = \frac{6+5}{4} = \frac{11}{4}$.

To calculate $h(-2)$, we substitute $x=-2$ into the function $h(x) = \frac{2x+5}{x+1}$:

$h(-2) = \frac{2(-2)+5}{-2+1} = \frac{-4+5}{-1} = \frac{1}{-1} = -1$.

(b) To find the antecedent(s) of 5, we need to solve the equation $h(x) = 5$ for $x$:

$\frac{2x+5}{x+1} = 5$

$2x+5 = 5(x+1)$

$2x+5 = 5x+5$

$0 = 3x$

$x = 0$

3. Final Answer

Part 1:

$g(x)$ is an even function.

$h(x)$ is an odd function.

Part 2:

(a) $h(3) = \frac{11}{4}$ and $h(-2) = -1$.

(b) The antecedent of 5 is

0.