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If \(f(x) = \frac{x^2}{x^4 - 1}\), 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. \(2f(x)\)
M15-17
A. \(f(x)\)
B. \(-f(x)\)
C. \(\frac{1}{f(x)}\)
D. \(-\frac{1}{f(x)}\)
E. \(2f(x)\)
M15-17
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07 Oct 2023, 10:31
Kudos
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If \(f(x) = \frac{x^2}{x^4 - 1}\), 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. \(2f(x)\)
\(f(\frac{1}{x})= \)
\(=\frac{(\frac{1}{x})^2}{(\frac{1}{x})^4 - 1} = \)
\(=\frac{\frac{1}{x^2} }{\frac{1}{x^4} - 1} = \)
\(=\frac{\frac{1}{x^2} }{\frac{1 - x^4}{x^4} } = \)
\(=\frac{1}{x^2}* \frac{x^4}{1 - x^4}= \)
\(=\frac{x^2}{1 - x^4} = \)
\(=-\frac{x^2}{x^4 - 1} = \)
\(=-f(x)\).
Answer: B
A. \(f(x)\)
B. \(-f(x)\)
C. \(\frac{1}{f(x)}\)
D. \(-\frac{1}{f(x)}\)
E. \(2f(x)\)
\(f(\frac{1}{x})= \)
\(=\frac{(\frac{1}{x})^2}{(\frac{1}{x})^4 - 1} = \)
\(=\frac{\frac{1}{x^2} }{\frac{1}{x^4} - 1} = \)
\(=\frac{\frac{1}{x^2} }{\frac{1 - x^4}{x^4} } = \)
\(=\frac{1}{x^2}* \frac{x^4}{1 - x^4}= \)
\(=\frac{x^2}{1 - x^4} = \)
\(=-\frac{x^2}{x^4 - 1} = \)
\(=-f(x)\).
Answer: B
General Discussion
16 Nov 2007, 08:21
Kudos
Bookmarks
bmwhype2
f(1/x) = (1/x)^2 / ((1/x)^4 - 1) = x^4/((x^2)* (1- x^4))
= x^2/(1-x^4)= - ( x^2/(x^4-1))
= -f(x)
I pick B.
What is OA?
