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For what values of \(x\) is \(x^{-1} < - 1\)?


Question

For what values of \(x\) is \(x^{-1} < - 1\)?

Options

A) 0 < x < 1

B) x 0

C) x > 1, x < 0

D) -1 < x < 0

The correct answer is D.

Explanation:

\(x^{-1}= < -1\)
\(\frac 1 x < -1\)
Multiply through by \(x^2\)
\(\frac 1 x \times \mathit{x}^2 < -1 \times x^2\)
\(x < -x^2\)
\(x+x^2 < 0\)
\(x\left(1+x\right) < 0\)
x = 0 or 1 + x = 0
x = 0 or -1
\(\begin{array}{cccc} Factor & x < -1 & - 1 < x < 0 & x > 0 \\ x & -ve & -ve & +ve \\ 1 + x & -ve & +ve & +ve \\ x(1 + x) & +ve & -ve & +ve \end{array}\)
Since \(x(1+x) < 0\) (i.e. negative) the solution of the inequality is \(-1 < x < 0\)

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Dicussion (1)

  • \(x^{-1}= < -1\)
    \(\frac 1 x < -1\)
    Multiply through by \(x^2\)
    \(\frac 1 x \times \mathit{x}^2 < -1 \times x^2\)
    \(x < -x^2\)
    \(x+x^2 < 0\)
    \(x\left(1+x\right) < 0\)
    x = 0 or 1 + x = 0
    x = 0 or -1
    \(\begin{array}{cccc} Factor & x < -1 & - 1 < x < 0 & x > 0 \\ x & -ve & -ve & +ve \\ 1 + x & -ve & +ve & +ve \\ x(1 + x) & +ve & -ve & +ve \end{array}\)
    Since \(x(1+x) < 0\) (i.e. negative) the solution of the inequality is \(-1 < x < 0\)

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