Complete the description of the piecewise function graphed below. Use interval notation to indicate intervals.

1st box = [-4, -3]

2nd box = (-3, 2]

3rd box = (2, 5]

For f(x) = -2 we see that x is between -4 to -3 with closed endpoints.

So the interval will be [-4, -3].

For f(x) = -5 we see that x is between -3 to 2 with closed endpoint at 2 and open endpoints at -3.

So the interval will be (-3, 2].

For f(x) = 5 we see that x is between 2 to 5 with open endpoint at 2 and closed endpoint at 5.

So the interval will be (2, 5].

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{ -2 if x ∈ -4 < x < -3

{ -5 if x ∈ -3 ≤ x < 2

[ 5 if x ∈ 2 ≤ x < 5    

Step-by-step explanation:

From the graph, we have;

{ -2 if x ∈ -4 < x < -3

For the next inequality line at y = -5, we have;

1) The inequality line, starts with an open circle on the left at x = -3, which is equivalent to less than or equal to symbol, ≤ as -3 ≤ x because the point x =-3 is the left boundary.  The inequality line extends to the point x = 2, where we have a closed circle (on the right), which is equivalent to a x ≤ 2 with the 2 on the right hand side because, the 2 is the right boundary of the inequality

2) The inequality can be summarized as starting from x = -3 extends to x = 2

3) As stated in point 1) the point x = 3, we have -3 ≤ x and at point x = 2, we have x ≤ 3 which can be combined as  -3 ≤ x < 2, to give;

{ -5 if x ∈ -3 ≤ x < 2

For the next inequality line at y = 5, we have;

1) The inequality line, starts with an open circle on the left at x = 2, which is equivalent to less than or equal to symbol, ≤ as 2 ≤ x because the point x =2 is the left boundary.  The inequality line extends to the right to reach point x = 5, where we have a closed circle (on the right), which is equivalent to a x ≤ 5 inequality, with the 5 on the right hand side because, the 5 is the right boundary of the inequality

2) The inequality can be summarized as starting from x = 2 extends to x = 5

3) As stated in point 1) the point x = 2, we have 2 ≤ x and at point x = 5, we have x ≤ 5 which can be combined as  2 ≤ x < 5, to give;

[ 5 if x ∈ 2 ≤ x < 5.

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)