Question

use the graph to determine the domain and range of the piecewise defined function. domain: -6≤x<0 or 0<x≤2; -6≤x≤0 or 0≤x≤2; 1<x≤6; 2<x≤6

Explanation:

Step1: Analyze the left part of the graph

The left - hand segment of the piece - wise function starts at \(x=-6\) (closed dot, so \(x = - 6\) is included) and goes up to \(x = 0\) (wait, looking at the graph, the left part of the piece - wise function: the first segment (the line with the peak) has a starting point at \(x=-6\) (closed circle) and then the other part? Wait, no, let's re - examine. Wait, the x - axis: the leftmost point of the graph is at \(x=-6\) (closed dot) and then the graph extends. Wait, the options are about the domain. Let's check the x - values covered. The left part of the piece - wise function (the polygonal part) has \(x\) from \(-6\) to \(0\) (inclusive) and then from \(0\) to \(2\)? Wait, no, wait the options: Let's check the x - coordinates of the endpoints. The left - most point is at \(x=-6\) (closed dot), then there is a point at \(x = 0\) (wait, the graph: the first part (the line with the peak) and then the other part. Wait, the domain is the set of all x - values for which the function is defined. Looking at the graph, the left segment (the line with the peak) starts at \(x=-6\) (closed circle, so \(x=-6\) is included) and goes to \(x = 0\)? Wait, no, the options: the first option is \(-6\leq x\lt0\) or \(0\lt x\leq2\), the second is \(-6\leq x\leq0\) or \(0\leq x\leq2\), the third is \(1\lt x\leq6\), the fourth is \(2\lt x\leq6\). Wait, maybe I misread the graph. Wait, the x - axis: the left part of the graph (the orange line with the peak) has a starting point at \(x=-6\) (closed dot) and then the other part (the curve) starts? Wait, no, the graph has two parts: one is a polygonal part (a triangle - like part) and one is a curve. Wait, the polygonal part (the line segments) has x - values from \(-6\) to \(0\) (inclusive) and from \(0\) to \(2\)? Wait, no, the closed dots: at \(x=-6\), at \(x = 0\) (maybe a closed dot?), and at \(x = 2\)? Wait, the second option is \(-6\leq x\leq0\) or \(0\leq x\leq2\). Wait, but if we consider that at \(x = 0\), the function is defined (since maybe there is a closed dot at \(x = 0\) for both parts? Wait, no, maybe the left part (the line) goes from \(x=-6\) to \(x = 0\) (inclusive) and the other part (the curve) starts? Wait, no, let's think again. The domain is the set of x - values. The left - most point is \(x=-6\) (closed), then the graph covers x from \(-6\) to \(0\) (inclusive) and then from \(0\) to \(2\) (inclusive). So the domain is \(-6\leq x\leq0\) or \(0\leq x\leq2\) (which is the same as \(-6\leq x\leq2\) because the union of \([-6,0]\) and \([0,2]\) is \([-6,2]\)). But among the options, the second option is \(-6\leq x\leq0\) or \(0\leq x\leq2\). Wait, but maybe the graph has a closed dot at \(x=-6\), a closed dot at \(x = 0\) (for the first part) and a closed dot at \(x = 2\) (for the second part of the polygonal part). Wait, the other options: the third and fourth are about \(x\gt1\) or \(x\gt2\), which don't match the left - most point at \(x=-6\). So the correct domain should be \(-6\leq x\leq0\) or \(0\leq x\leq2\) (the second option).

Step2: Eliminate other options

• Option 3 (\(1\lt x\leq6\)) and Option 4 (\(2\lt x\leq6\)) can be eliminated because the left - most point of the graph is at \(x=-6\), so the domain must include \(x=-6\), which these options do not.
• Option 1 (\(-6\leq x\lt0\) or \(0\lt x\leq2\)) is incorrect because if there is a closed dot at \(x = 0\) (which is likely, as the graph seems to be defined at \(x = 0\) for both parts, or at least the function is defined at \(x = 0\)), then \(x = 0\) should be included in the…

Answer:

\(-6\leq x\leq0\) or \(0\leq x\leq2\) (the second option, so the answer is the option with \(-6\leq x\leq0\) or \(0\leq x\leq2\))