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16 Sep 2014, 01:09
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If \(f(x) = \frac{x^4 - 1}{x^2}\), what is \(f(\frac{1}{x})\) in terms of \(f(x)\)?
A. \(f(x)\)
B. \(-f(x)\)
C. \(\frac{1}{f(x)}\)
D. \(-\frac{1}{f(x)}\)
E. \(2f(x)\)
A. \(f(x)\)
B. \(-f(x)\)
C. \(\frac{1}{f(x)}\)
D. \(-\frac{1}{f(x)}\)
E. \(2f(x)\)
16 Sep 2014, 01:09
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Bookmarks
Official Solution:
If \(f(x) = \frac{x^4 - 1}{x^2}\), what is \(f(\frac{1}{x})\) in terms of \(f(x)\)?
A. \(f(x)\)
B. \(-f(x)\)
C. \(\frac{1}{f(x)}\)
D. \(-\frac{1}{f(x)}\)
E. \(2f(x)\)
\(f(\frac{1}{x}) = \)
\(=\frac{\frac{1}{x^4} - 1}{\frac{1}{x^2} } = \)
\(=(\frac{1}{x^4} - 1)*x^2 = \)
\(=\frac{1}{x^2} - x^2 = \)
\(=\frac{1 - x^4}{x^2} = \)
\(=-\frac{x^4 - 1}{x^2} = \)
\(=-f(x)\).
Answer: B
If \(f(x) = \frac{x^4 - 1}{x^2}\), what is \(f(\frac{1}{x})\) in terms of \(f(x)\)?
A. \(f(x)\)
B. \(-f(x)\)
C. \(\frac{1}{f(x)}\)
D. \(-\frac{1}{f(x)}\)
E. \(2f(x)\)
\(f(\frac{1}{x}) = \)
\(=\frac{\frac{1}{x^4} - 1}{\frac{1}{x^2} } = \)
\(=(\frac{1}{x^4} - 1)*x^2 = \)
\(=\frac{1}{x^2} - x^2 = \)
\(=\frac{1 - x^4}{x^2} = \)
\(=-\frac{x^4 - 1}{x^2} = \)
\(=-f(x)\).
Answer: B
18 Oct 2017, 23:06
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\(\)
Just plug in \(\frac{1}{2}\) and \(2\), then compare the answers.
For \(f(\frac{1}{2})\), you had get \((\frac{1}{16} -1) / \frac{1}{4} = \frac{-15}{4}\)
and for \(f(2)\), you had get \(\frac{15}{4}\) i.e \(\frac{(16-1)}{4}\)
Thus \(f(\frac{1}{x})\) = \(-f(x)\), which is B.
alternatively, algebraically, \(f(x) = \frac{x^4 - 1}{x^2}\) = \(x^2 - \frac{1}{x^2}\)
One can quickly see the pattern and know that the function is negatively symmetrical for the reciprocal i.e let \(x^2 =\frac{1}{2}\), then \(x^2 - \frac{1}{x^2}\)= negative and positive same value if \(x^2 =2\). Again, B.
Bunuel
Just plug in \(\frac{1}{2}\) and \(2\), then compare the answers.
For \(f(\frac{1}{2})\), you had get \((\frac{1}{16} -1) / \frac{1}{4} = \frac{-15}{4}\)
and for \(f(2)\), you had get \(\frac{15}{4}\) i.e \(\frac{(16-1)}{4}\)
Thus \(f(\frac{1}{x})\) = \(-f(x)\), which is B.
alternatively, algebraically, \(f(x) = \frac{x^4 - 1}{x^2}\) = \(x^2 - \frac{1}{x^2}\)
One can quickly see the pattern and know that the function is negatively symmetrical for the reciprocal i.e let \(x^2 =\frac{1}{2}\), then \(x^2 - \frac{1}{x^2}\)= negative and positive same value if \(x^2 =2\). Again, B.
