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Assume the function
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Updated on: 30 Dec 2017, 10:38
Updated on: 30 Dec 2017, 10:38
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14% (00:35) correct
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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
Indicate all possible values.
A)6
B)5
C)4
D)3
E)2
F)1
Show: :: OA
A,B,C,D,E
ShowHide Answer
Official Answer
A,B,C,D,E
Originally posted by Sonalika42 on 23 Sep 2016, 01:23.
Last edited by Carcass on 30 Dec 2017, 10:38, edited 2 times in total.
Last edited by Carcass on 30 Dec 2017, 10:38, edited 2 times in total.
Edited by Carcass
Re: Assume the function
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24 Sep 2016, 06:46
24 Sep 2016, 06:46
Expert Reply
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
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.??
Re: Assume the function
[#permalink]
12 Oct 2016, 15:09
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
Indicate all possible values.
A)6
B)5
C)4
D)3
E)2
F)1
