Given f(x) = sqrt(x−1) and g(x) = (x)/(x+2), find f g(x) and g f(x), stating their domains.

Substitute $g(x)$ into $f(x)$:

Simplify the expression inside the square root:

So,

For $f \circ g(x)$ to be defined:

1. $x$ must be in the domain of $g$, so $x \ne -2$.
2. $g(x)$ must be in the domain of $f$, so $g(x) \ge 1$. $$\frac{x}{x+2} \ge 1$$ $$\frac{x}{x+2} - 1 \ge 0$$ $$\frac{x - (x+2)}{x+2} \ge 0$$ $$\frac{-2}{x+2} \ge 0$$ For this inequality to hold, since the numerator is negative, the denominator must be negative (and cannot be zero). $$x+2 < 0 \implies x < -2$$
3. The expression inside the square root must be non-negative: $\frac{-2}{x+2} \ge 0$. This leads to the same condition $x < -2$. Combining these conditions, the domain of $f \circ g(x)$ is $\{x \in \mathbb{R} \mid x < -2\}$.

Substitute $f(x)$ into $g(x)$:

For $g \circ f(x)$ to be defined:

1. $x$ must be in the domain of $f$, so $x \ge 1$.
2. $f(x)$ must be in the domain of $g$, so $f(x) \ne -2$. $\sqrt{x-1} \ne -2$. Since $\sqrt{x-1} \ge 0$ for $x \ge 1$, this condition is always satisfied.
3. The denominator of $g(f(x))$ cannot be zero: $\sqrt{x-1}+2 \ne 0$. Since $\sqrt{x-1} \ge 0$, $\sqrt{x-1}+2 \ge 2$. Thus, the denominator is never zero. This condition is always satisfied for $x \ge 1$. Combining these conditions, the domain of $g \circ f(x)$ is $\{x \in \mathbb{R} \mid x \ge 1\}$.

A function $g: X \to Y$ is surjective if for every element $y$ in the codomain $Y$, there exists at least one element $x$ in the domain $X$ such that $g(x) = y$.
For $g(x) = \frac{x}{x+2}$, the domain is $X = \mathbb{R} \setminus \{-2\}$ and the codomain is $Y = \mathbb{R}$.

Let $y = g(x)$.

Multiply both sides by $(x+2)$:

Rearrange to solve for $x$:

If $1 - y = 0$, i.e., $y = 1$, the equation becomes $2(1) = x(0)$, which simplifies to $2 = 0$. This is a contradiction.
This means that there is no value of $x$ in the domain of $g$ for which $g(x) = 1$.
Since 1 is an element of the codomain $\mathbb{R}$, but it has no corresponding $x$ in the domain, the function $g$ is not surjective.

The condition $a+b = 2n, n \in \mathbb{N}$ means that $a+b$ must be an even positive integer. Since $a, b \in \{1, 2, 3, 4, 5\}$, the smallest sum is $1+1=2$ and the largest is $5+5=10$. All these sums are positive.
Therefore, the condition simplifies to $a+