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May 22nd, 2019, 02:37 AM   #11
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So range of y should be (-0,1] -1/sqrt(2) Right ?
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May 22nd, 2019, 02:54 AM   #12
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Select choice (c) (as given in the question) as the correct answer.
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May 22nd, 2019, 01:45 PM   #13
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Originally Posted by skeeter View Post
because $x = 9$ forms a removable discontinuity for $y = \sin[2 \cdot f(x)]$
(see original equation for $f$) ... mea culpa.

domain should be $x \in [-7, 9) \cup (9, \infty)$

good catch by skipjack.
I had looked at the question after the discontinuity had been removed.
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