The problem consists of two sub-problems: (4) Determine whether the function $f(x) = -x + 5$ defined on the interval $0 \le x < 2$ is odd, even, neither, or none of the above. (5) Determine which of the following functions is periodic: (a) $x^2$, (b) $\frac{2x+1}{x^2}$, (c) $\sin(3x)$, (d) none of the above.

1. Problem Description

The problem consists of two sub-problems:

(4) Determine whether the function $f(x) = -x + 5$ defined on the interval $0 \le x < 2$ is odd, even, neither, or none of the above.

(5) Determine which of the following functions is periodic: (a) $x^2$, (b) $\frac{2x+1}{x^2}$, (c) $\sin(3x)$, (d) none of the above.

2. Solution Steps

(4)

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 $f$.

The domain of $f(x)$ is $0 \le x < 2$. Since the domain does not contain any negative values, we cannot directly check if the function is even or odd.

However, we can check by assuming it's an even or odd function, then verifying if the definition holds for some value of $x$.

Suppose $f(x)$ is even. Then $f(-x) = f(x)$ for all $x$. But $f(-x)$ is not defined as the domain is $0 \le x < 2$, so it's not even.

Suppose $f(x)$ is odd. Then $f(-x) = -f(x)$. Again, $f(-x)$ is not defined as the domain is $0 \le x < 2$, so it's not odd.

Also, $f(1) = -1+5 = 4$. If $f(x)$ were even, we would need $f(-1) = 4$. If $f(x)$ were odd, we would need $f(-1) = -4$. Neither is possible.

Thus, the function is neither odd nor even.

(5)

A function $f(x)$ is periodic if there exists a non-zero constant $T$ such that $f(x+T) = f(x)$ for all $x$ in the domain of $f$.

(a) $f(x) = x^2$. $f(x+T) = (x+T)^2 = x^2 + 2xT + T^2$. If $f(x+T) = f(x)$, then $x^2 + 2xT + T^2 = x^2$, which implies $2xT + T^2 = 0$. Since this must be true for all $x$, we require $T=0$. Therefore, $x^2$ is not periodic.

(b) $f(x) = \frac{2x+1}{x^2}$. $f(x+T) = \frac{2(x+T)+1}{(x+T)^2} = \frac{2x+2T+1}{x^2+2xT+T^2}$. It's clear that $f(x+T) \neq f(x)$ for any $T \neq 0$. Therefore, $\frac{2x+1}{x^2}$ is not periodic.

(c) $f(x) = \sin(3x)$. The general form of a sine function is $\sin(ax)$, which has a period of $\frac{2\pi}{|a|}$. Here, $a=3$, so the period is $\frac{2\pi}{3}$. Therefore, $\sin(3x)$ is periodic.

Therefore, the periodic function is $\sin(3x)$.

3. Final Answer

4. Neither odd nor even

5. sin 3x