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Bunuel
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If f(x) = (x +1)^(1/2), and if the domain of x is the set {3,8,15} 11 Dec 2020, 11:00
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If \(f(x) = \sqrt{x +1}\), and if the domain of x is the set {3,8,15}, then which of the following sets indicates all possible values of f(x) ?

(A) {–4, –3, –2, 2, 3, 4}
(B) {2, 3, 4}
(C) {4, 9, 16}
(D) {3, 8, 15}
(E) {all real numbers}
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B
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If f(x) = (x +1)^(1/2), and if the domain of x is the set {3,8,15} 11 Dec 2020, 20:32
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If \(f(x) = \sqrt{x +1}\), and if the domain of x is the set {3,8,15}, then which of the following sets indicates all possible values of f(x) ?

(A) {–4, –3, –2, 2, 3, 4}
(B) {2, 3, 4}
(C) {4, 9, 16}
(D) {3, 8, 15}
(E) {all real numbers}
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1) x=3......\(f(x)=\sqrt{x+1}=\sqrt{4}=2\)...Square roots are always positive
2) x=8......\(f(x)=\sqrt{x+1}=\sqrt{9}=3\)
3) x=15......\(f(x)=\sqrt{x+1}=\sqrt{16}=4\)

B
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If f(x) = (x +1)^(1/2), and if the domain of x is the set {3,8,15} 14 Dec 2020, 04:30
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Bunuel
If \(f(x) = \sqrt{x +1}\), and if the domain of x is the set {3,8,15}, then which of the following sets indicates all possible values of f(x) ?

(A) {–4, –3, –2, 2, 3, 4}
(B) {2, 3, 4}
(C) {4, 9, 16}
(D) {3, 8, 15}
(E) {all real numbers}
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Solution:

When x = 3, f(3) = √(3 + 1) = √4 = 2.

When x = 8, f(8) = √(8 + 1) = √9 = 3.

When x = 15, f(15) = √(15 + 1) = √16 = 4.

We see that when the domain is {3, 8, 15}, the possible values of f(x) (i.e, the range) is {2, 3, 4}.

Answer: B
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If f(x) = (x +1)^(1/2), and if the domain of x is the set {3,8,15} 16 Dec 2020, 21:35
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I want to know something. The value of square root of an integer comes with a positive and also a negative of the same number. So can't the positive 2,3,4 have also negative 2,3, 4?

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If f(x) = (x +1)^(1/2), and if the domain of x is the set {3,8,15} 12 Jan 2024, 10:36
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square root doesn't contain -ve part
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If f(x) = (x +1)^(1/2), and if the domain of x is the set {3,8,15} 05 Sep 2024, 08:31
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I want to know something. The value of square root of an integer comes with a positive and also a negative of the same number. So can't the positive 2,3,4 have also negative 2,3, 4?

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­I agree. Can some Expert please shed some light on this? Bunuel GMATNinja

Since 2^2 = 4 and (-2)^2 = 4, isn't square root of 4 = +2,-2?
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If f(x) = (x +1)^(1/2), and if the domain of x is the set {3,8,15} 05 Sep 2024, 08:34
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Hea234ven
I want to know something. The value of square root of an integer comes with a positive and also a negative of the same number. So can't the positive 2,3,4 have also negative 2,3, 4?

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­I agree. Can some Expert please shed some light on this? Bunuel GMATNinja

Since 2^2 = 4 and (-2)^2 = 4, isn't square root of 4 = +2,-2?
Show more
­


Even roots cannot give negative result.

\(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;
Similarly \(\sqrt{\frac{1}{16}} = \frac{1}{4}\), NOT +1/4 or -1/4.


Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).­
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If f(x) = (x +1)^(1/2), and if the domain of x is the set {3,8,15} 05 Sep 2024, 09:00
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Bunuel

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Hea234ven
I want to know something. The value of square root of an integer comes with a positive and also a negative of the same number. So can't the positive 2,3,4 have also negative 2,3, 4?

Posted from my mobile device
­I agree. Can some Expert please shed some light on this? Bunuel GMATNinja

Since 2^2 = 4 and (-2)^2 = 4, isn't square root of 4 = +2,-2?
­


Even roots cannot give negative result.

\(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;
Similarly \(\sqrt{\frac{1}{16}} = \frac{1}{4}\), NOT +1/4 or -1/4.


Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).­
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­Thank you so much! It makes sense now.
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If f(x) = (x +1)^(1/2), and if the domain of x is the set {3,8,15} 08 Sep 2025, 11:07
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