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May 20th, 2019, 02:12 AM   #1
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Function :- Range Problem

If f(x) = pi( ( (sqrt(x-7)) -4)/((x-9)) then the range of y= sin( 2f(x)) is
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May 20th, 2019, 06:00 AM   #2
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If f(x) = pi( ( (sqrt(x-7)) -4)/((x-9)) . . .
This doesn't make sense - the parentheses aren't paired properly.
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May 20th, 2019, 08:59 AM   #3
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Please refer to Question # 2 of the attachment
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May 20th, 2019, 02:41 PM   #4
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Try to express your question more clearly - use LaTex.
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May 20th, 2019, 02:54 PM   #5
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Please refer to Question # 2 of the attachment
Did you mean Question # 3?
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May 20th, 2019, 03:31 PM   #6
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note ...

$\pi \left[\dfrac{\sqrt{x+7}-4}{x-9}\right] = \dfrac{\pi}{\sqrt{x+7}+4}$

$y = \sin\left(\dfrac{2\pi}{\sqrt{x+7}+4}\right)$

$x \in [-7,\infty) \implies y \in \{?\}$
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May 21st, 2019, 01:24 AM   #7
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Many Thanks. Got it . So range of y should be (-0,1] Right?
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May 21st, 2019, 04:06 AM   #8
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No. As x = 9 isn't in the domain, 1/√2 isn't in the range.
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May 21st, 2019, 01:43 PM   #9
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No. As x = 9 isn't in the domain, 1/√2 isn't in the range.
If the domain is$[-7,\infty)$, why is $x=9$ not in the domain?
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May 21st, 2019, 04:40 PM   #10
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If the domain is$[-7,\infty)$, why is $x=9$ not in the domain?
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.
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