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If g(x) = , for which of the following x values is g(x) unde
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30 Jul 2018, 10:19
30 Jul 2018, 10:19
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71% (00:42) correct
28% (01:14) wrong based on 64 sessions
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If \(g(x) = \frac{x^2(4x+9)}{(3x-3)(x+2)}\), for which of the following x values is g(x) undefined?
Indicate all such values of x.
A. \(-\frac{9}{4}\)
B. \(–2\)
C. \(0\)
D. \(1\)
E. \(2\)
F. \(\frac{9}{4}\)
Indicate all such values of x.
A. \(-\frac{9}{4}\)
B. \(–2\)
C. \(0\)
D. \(1\)
E. \(2\)
F. \(\frac{9}{4}\)
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Re: If g(x) = , for which of the following x values is g(x) unde
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03 Aug 2018, 12:27
03 Aug 2018, 12:27
1
sandy wrote:
If \(g(x) = \frac{x^2(4x+9)}{(3x-3)(x+2)}\), for which of the following x values is g(x) undefined?
Indicate all such values of x.
A. \(\frac{-9}{4}\)
B. \(–2\)
C. \(0\)
D. \(1\)
E. \(2\)
F. \(\frac{9}{4}\)
Indicate all such values of x.
A. \(\frac{-9}{4}\)
B. \(–2\)
C. \(0\)
D. \(1\)
E. \(2\)
F. \(\frac{9}{4}\)
A rational expression (aka fraction) is considered undefined when the denominator is zero
In this question the denominator is (3x-3)(x+2)
So, the denominator will equal zero when (3x-3) = 0 or (x+2) = 0
Let's examine each case separately.
If (3x-3) = 0, the x = 1
If (x+2) = 0, the x = -2
So, when x = 1, the fraction is undefined.
Likewise, when x = -2, the fraction is undefined.
Answer: B, D
Cheers,
Brent
Re: If g(x) = , for which of the following x values is g(x) unde
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11 Aug 2018, 16:38
11 Aug 2018, 16:38
2
Expert Reply
Explanation
The term “undefined” refers to the circumstance when the solution is not a real number —for example, division by 0 is considered “undefined.” There aren’t many circumstances that result in an undefined answer.
Essentially, the GRE considers two main cases: the square root of a negative integer and division by 0. There are no square roots in this problem, but it’s possible that 0 could end
up on the denominator of the fraction. Set each of the terms in the denominator equal to 0:
\(3x - 3 = 0\)
\(3x = 3\)
\(x = 1\)
\(x + 2 = 0\)
\(x = -2\)
Thus, if x = 1 or x = –2, then the denominator would be 0, making g(x) undefined. All other values are acceptable.
The term “undefined” refers to the circumstance when the solution is not a real number —for example, division by 0 is considered “undefined.” There aren’t many circumstances that result in an undefined answer.
Essentially, the GRE considers two main cases: the square root of a negative integer and division by 0. There are no square roots in this problem, but it’s possible that 0 could end
up on the denominator of the fraction. Set each of the terms in the denominator equal to 0:
\(3x - 3 = 0\)
\(3x = 3\)
\(x = 1\)
\(x + 2 = 0\)
\(x = -2\)
Thus, if x = 1 or x = –2, then the denominator would be 0, making g(x) undefined. All other values are acceptable.
