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y=x^3 and y=(x+2)^3+1

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The domains and ranges of both equations are all real numbers, since there are no limits on what x can be in either of the problem, and y can also be any number infinitely great or infinitely small depending on the x value.
 
To find where the graphs are increasing or decreasing, you must take the first derivative of each graph and set those equal to 0, finding the relative maximums, minimums, or turning points of each graph.
 
The derivative of y=x^3 is y'=3x^2. y=0 when x=0, so that means that the graph is horizontal at x=0, and neither increasing nor decreasing. Since we need to find where it is increasing or decreasing, we need to look at the graph on both sides of x=0. Let's look at x=-1 and x=1. When x=-1, 3*(-1)^2 = 3. Since 3 is a positive number, the graph is INCREASING to the left of x=0. when x=1, 3*(1)^2 = 3. For the same reason as before, the graph is INCREASING to the right of x=0. Therefore, the graph y=x^3 is increasing from (-∞,0) and (0,∞), which can be written as (-∞,0) U (0,∞). It MUST be written this way, as the graph is NOT increasing specifically at the point x=0.
 
The graph y=(x+2)^3+1 is similar to the graph y=x^3, except that it is moved up 1 and to the left 2. This means that all you have to do to find where it is increasing and decreasing is subtract 2 from the values we found for the previous graph. (-∞-2, 0-2) U (0-2,∞-2). Since infinity is infinitely large, no amount of adding or subtracting from it will ever change it's value, therefore the new values for where the graph is increasing are (-∞,-2) U (-2,∞).
 
In conclusion:
The domain and range for both equations is ALL REAL NUMBERS, also written as (-∞,∞)
The graph y=x^3 is increasing (-∞,0) U (0,∞) and decreasing NOWHERE.
The graph y=(x+2)^3+1 is increasing (-∞,-2) U (-2,∞) and decreasing NOWHERE.
 
Hope this helps!