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Question 902504: Consider the graph:
http://i.imgur.com/vEeIKTv.png?1
I think this is a very simple problem and I am just missing something very obvious but how would this be solved? How should I understand the graph provided.
Thank you
My objective is to choose the right options from below
(a) Determine the interval(s) on which the function is increasing. (Select all that apply.)
[-2, -1)
[-2, 1]
[-2, 2]
[-1, 1]
(1, 2]
[-1, 2]
(b) Determine the interval(s) on which the function is decreasing. (Select all that apply.)
[-2, -1)
[-2, 1]
[-2, 2]
[-1, 1]
(1, 2]
[-1, 2]
Thank You in advance
Answer by Stitch(470)   (Show Source):
You can put this solution on YOUR website! 
Read the graph from left to right. When the line is in an upward direction the function is increasing. When the line is going in a downward direction, the function is decreasing. It is also important to note the open circles on the graph (the purple ones). An open circle indicates that particular value is not included in the graph but all the numbers up to it are. (See the description below about brackets and parenthesis.
For question A:
The function is increasing in the interval of [-1,1]
[-1,1] indicates all real numbers between and including -1 to 1
For question B:
The function is decreasing in the intervals of [-2,-1) & (1,2]
For example: [-2,-1) indicates all real numbers between -2 and -1, but does not include -1.
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I copied this description of bracket and parenthesis from Yahoo
"Where a and b are the lower and upper bounds of the set respectively:
( a , b ) refers to an open interval, a set in which all real numbers between a and b excluding a and y are included.
[ a , b ] refers to an closed interval, a set containing all real numbers between and including a and b.
[ a , b) and ( a , b ] refer to half-open intervals.
The half-open interval [ a , b ) refers to a set which includes all real numbers between a and b as well as a but not b.
The half-open interval ( a , b ] refers to a set which includes all real numbers between a and b as well as b but not a."
-Srinivas V
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