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[HoldingOn]

1. How would I find the domain of 1/1-sin^(2)x?
2. An inequality which has the same solution as |x+2| + |x-3|=5?

Thanks in advance.

 

[darkk_blu]

For 1 :



From here, you can either visualise the graph of and look at its domain or visualise the graph its x intercepts will be your restrictions and determine your domain like that.

 

[fan96]

1.



Therefore, this expression is undefined when .

2. You can either take cases, or use the definition .

(you will end up with )

 

[HoldingOn]

1.



Therefore, this expression is undefined when .

2. You can either take cases, or use the definition .

(you will end up with )

How exactly would you apply the definition? Thanks

 

[fan96]

How exactly would you apply the definition? Thanks

Actually forget what I said, I misread the question.

The easiest way to solve this would probably be graphing.

You could:

i) graph and , add their graphs and check where it equals 5, or

ii) square both sides (which doesn't introduce new solutions in this case because both sides are positive) and rearrange to get , which should be much simpler to graph.

 

[InteGrand]

2. An inequality which has the same solution as |x+2| + |x-3|=5?

Thanks in advance.

If you don't want to graph stuff, you can also do it algebraically in cases.

The cases to consider are x ≤ -2, -2 < x < 3, and x ≥ 3 (because these are the places where the inputs to the absolute value change sign).

E.g. if x ≤ -2, then x + 2 ≤ 0, so |x+2| = -(x+2) = -x-2, and x - 3 ≤ -5 < 0, so |x-3| = -(x-3) = 3-x. Hence the equation to solve becomes -x - 2 + 3 - x = 5, i.e. -2x + 1 = 5. This gives 2x = -4, i.e. x = -2. This satisfies the condition of this case (x ≤ -2). So one solution is x = -2 (you can and should also verify this by plugging x = -2 into the given equation).

Repeat similar analysis for the remaining two cases to finish the problem.