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Assume the function

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Manager
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Joined: 12 Jan 2016
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Assume the function [#permalink] New post 23 Sep 2016, 01:23
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Question Stats:

0% (00:00) correct 100% (00:36) wrong based on 2 sessions
Assume the function \(f(x)=(x - 5)^2+sqrt(x+3)+(\frac{5}{(x+2)})\) For which of the values \(f(x)\) is defined?

Indicate all possible values.

A)6
B)5
C)4
D)3
E)2
F)1

[Reveal] Spoiler: OA
A,B,C,D,E
[Reveal] Spoiler: OA

Last edited by Carcass on 30 Dec 2017, 10:38, edited 2 times in total.
Edited by Carcass
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GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Re: Assume the function [#permalink] New post 24 Sep 2016, 06:46
Expert's post
Sonalika42 wrote:
Assume the function \(f(x)=(x - 5)^2+sqrt(x+3)+(\frac{5}{(x+2)})\) For which of the values \(f(x)\) is defined?

Indicate all possible values.

A)6
B)5
C)4
D)3
E)2
F)1



I am assuming you mean f(x) is a real valued function and we need to select all options where f(x) is a real number.

In the function there is a square root term, \(sqrt(x+3)\) and the term \(\frac{5}{(x+2)}\). That can yield non real values/ undefined values.

\(\frac{5}{(x+2)}\) is undefined for x=-2. For x= -2 \(\frac{5}{(x+2)}\) becomes \(\frac{5}{(-2+2)}\) or \(\frac{5}{0}\). Which is undefined.

\(sqrt(x+3)\) is not real is \(x+3 < 0\), or \(x< -3\).

So as long as x is greater than or equal to -3. and x is not equal to -2, \(f(x)\) is a real number.

Hence all the options A, B, C, D, E are valid.

PS: Could you please share the page number of his question.??
_________________

Sandy
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Intern
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Joined: 21 Oct 2015
Posts: 28
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Kudos [?]: 10 [0], given: 3

Re: Assume the function [#permalink] New post 12 Oct 2016, 15:09
can you post answer of this question?


Sonalika42 wrote:
Assume the function \(f(x)=(x - 5)^2+sqrt(x+3)+(\frac{5}{(x+2)})\) For which of the values \(f(x)\) is defined?

Indicate all possible values.

A)6
B)5
C)4
D)3
E)2
F)1
Intern
Intern
Joined: 15 Dec 2016
Posts: 18
Followers: 0

Kudos [?]: 6 [0], given: 3

Re: Assume the function [#permalink] New post 18 Dec 2016, 15:34
I think F is also a correct answer.
Re: Assume the function   [#permalink] 18 Dec 2016, 15:34
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Assume the function

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