SOLUTION: How do you determine the range of this function: y = 2 + 2sec(2x)

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Question 628076: How do you determine the range of this function:
y = 2 + 2sec(2x)
Found 2 solutions by solver91311, Edwin McCravy:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Just looking at the secant function, and it has no upper bound. Since the absolute value of secant can be no less than 1,

Range:

John

My calculator said it, I believe it, that settles it
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Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
y = 2 + 2sec(2x)

The upper part of the range will be when the secant has the smallest
positive value up to infinity.

The smallest positive value of the secant is 1

So the minimum of the upper part of the range of

y = 2 + 2sec(2x) is 2 + 2(1) = 2 + 2 = 4

So the upper part of the range is [4, infinity)

The lower part of the range will be from negative infinity
up to when the secant has the largest negative value.

The largest negative value of the secant is -1

So the maximum of the lower part of the range of

y = 2 + 2sec(2x) is 2 + 2(-1) = 2 - 2 = 0

So the lower part of the range is (-infinity, 0].

Therefore the range is (-infinity, 0] U [4, infinity)

Edwin