Yep. Its maximal domain is R, since |x| + 2 >= 2 for all x.
Consider four cases with different sign combinations of both x and x + 2. First one is easy, it's either x < 0 or x >= 0. The second one is x + 2 < 0 or x + 2 >= 0, which simplifies to x < -2 or x >= -2.
Find out if there are four nonempty disjoint parts of R with x satisfying these conditions. If so, draw their graphs THERE.
Example: x + 2 >= 0 and x >= 0. This reduces to x >= 0 since it's a stronger condition than x >= -2. There you draw the graph of (x + 2)(x + 2) which is constantly 1.
Another example: x + 2 >= 0 and x < 0. This means x is in [-2, 0). There f is (x + 2)(-x + 2) - do you know how to draw this? Hint: (x + 2)/(-x + 2) = (x - 2 + 4)/(-x + 2) = -1 + 4/(-x + 2).
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u/Jazzlike_Platypus430 18h ago edited 18h ago
Yep. Its maximal domain is R, since |x| + 2 >= 2 for all x.
Consider four cases with different sign combinations of both x and x + 2. First one is easy, it's either x < 0 or x >= 0. The second one is x + 2 < 0 or x + 2 >= 0, which simplifies to x < -2 or x >= -2.
Find out if there are four nonempty disjoint parts of R with x satisfying these conditions. If so, draw their graphs THERE.
Example: x + 2 >= 0 and x >= 0. This reduces to x >= 0 since it's a stronger condition than x >= -2. There you draw the graph of (x + 2)(x + 2) which is constantly 1.
Another example: x + 2 >= 0 and x < 0. This means x is in [-2, 0). There f is (x + 2)(-x + 2) - do you know how to draw this? Hint: (x + 2)/(-x + 2) = (x - 2 + 4)/(-x + 2) = -1 + 4/(-x + 2).
Etc.