We are given the function $f(x) = |x-5| - 1$. We need to determine if the function is even, odd, or neither, state the range of $f(x)$, and sketch the graph of $f(x)$.

1. Problem Description

We are given the function $f(x) = |x-5| - 1$. We need to determine if the function is even, odd, or neither, state the range of $f(x)$, and sketch the graph of $f(x)$.

2. Solution Steps

(i) To determine if a function is even, we check if $f(-x) = f(x)$. To determine if a function is odd, we check if $f(-x) = -f(x)$.

$f(-x) = |-x - 5| - 1 = |-(x+5)| - 1 = |x+5| - 1$.

Since $f(x) = |x-5| - 1$, and $f(-x) = |x+5| - 1$, we see that $f(x) \neq f(-x)$, so the function is not even.

Also, $-f(x) = -(|x-5| - 1) = -|x-5| + 1$. Since $f(-x) = |x+5| - 1$ and $-f(x) = -|x-5| + 1$, we see that $f(-x) \neq -f(x)$, so the function is not odd.

Therefore, the function is neither even nor odd.

(ii) The absolute value function $|x-5|$ is always non-negative, so $|x-5| \geq 0$ for all $x$.

Then, $|x-5| - 1 \geq 0 - 1$, so $|x-5| - 1 \geq -1$.

The minimum value of $|x-5|$ is 0, which occurs when $x=5$. In that case, $f(5) = |5-5| - 1 = 0 - 1 = -1$.

As $x$ moves away from 5 in either direction, $|x-5|$ increases without bound.

Thus, the range of $f(x)$ is $[-1, \infty)$.

(iii) To sketch the graph, we start with the absolute value function $y = |x|$. Then, we consider $y = |x-5|$, which is the graph of $y = |x|$ shifted 5 units to the right. Finally, we consider $y = |x-5| - 1$, which is the graph of $y = |x-5|$ shifted down 1 unit.

The vertex of the graph is at $(5, -1)$. The graph is symmetric around $x=5$. When $x=0$, $f(0) = |0-5| - 1 = 5 - 1 = 4$.

The graph intersects the x-axis when $f(x) = 0$, which means $|x-5| - 1 = 0$, so $|x-5| = 1$.

This gives us $x-5 = 1$ or $x-5 = -1$. Therefore, $x = 6$ or $x = 4$. The x-intercepts are at (4,0) and (6,0).

3. Final Answer

(i) Neither

(ii) $[-1, \infty)$

(iii) The graph is a V-shape with vertex at $(5, -1)$. The x-intercepts are at $x=4$ and $x=6$.