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Chapter 7: Problem 4

If \(-1

Short Answer

Expert verified
The correct option is the third one: \(-1 < x < 0 < x^2\).

Step by step solution

01

Interpret the Given Information

The first step is to understand that \(x\) lies between 0 and \(-1\), which implies that \(x\) is a negative number with greater magnitude than \(-1\).
02

Check First Option: \(x^2 < -1 < x < 0\)

Now consider the inequality \(x^2<-1\). Since the square of a real number is always nonnegative, \(x^2\) will never be less than \(-1\), hence the first option is incorrect.
03

Check Second Option: \(-1 < x < x^2 < 0\)

Now observe that, within given interval, the inequality \(x < x^2\) always holds true because \(x^2\) will always be a positive number (greater than \(x\) which is negative), and this positive number is less than 0. Therefore, the second option is incorrect.
04

Check Third Option: \(-1 < x < 0 < x^2\)

For this inequality to hold, \(x^2\) should always be positive and greater than 0 in the given interval for \(x\). As we have noted, \(x^2\) will indeed be positive (greater than \(x\)), and hence this inequality holds true for all \(x\) in the given range. So, the third option is correct.

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Most popular questions from this chapter

If \(x<0\), which of the following is true? C \(x^3\) is positive. ( \(x^3\) is negative. \(x^2\) could be either positive or negative. I don't know.

If \(x^2=25\), which of the following is true? $$ \begin{aligned} & \bigcirc x=5 \\ & \bigotimes x=-5 \\ & \hdashline x=\pm 5 \\ & \text { idon't know. } \end{aligned} $$

Which of the following expressions is equivalent to \(\frac{12 !}{15 !} ?\) $$ \begin{aligned} & \text { } \frac{\frac{4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1}}{\frac{1}{15 \times 14 \times 13}} \\ & \frac{4}{5} \end{aligned} $$ I don't know.

Which of the following is equivalent to \(x^{\frac{3}{2}}\) ? C \(\sqrt{x^3}\) \(\frac{3}{2} x\) \(\frac{x^2}{x^2}\) I don't know.

Line 1 is defined by \(y=3 x+7\), and line 2 is perpendicular to line 1. What is the slope of line 2 ? \(\bigcirc-3\) O \(\frac{1}{3}\) \(\bigcirc-\frac{1}{3}\) O I don't know.
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