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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 $$-1 < x < 0$$

## 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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