Q. Solve the inequality \( \left| -x^{-2} \right| < 1 \).

Q: What does the notation -x^{-2} mean?

It means -x^{-2} = -\left(x^{-2}\right) = -1/x^{2} ; the exponent applies to x first, then the leading minus multiplies the result.

Q: How do you simplify \lvert -x^{-2} \rvert ?

\lvert -x^{-2} \rvert = \lvert -1/x^{2} \rvert = 1/x^{2} for x \neq 0 , since absolute value removes the negative sign..

Q: How do you solve the inequality |-x^{-2}| < 1 ?

Replace with 1/x^{2} < 1. Multiply by x^{2}>0 ( x \neq 0 ) to get 1 < x^{2}, so x^{2} > 1. Hence \lvert x \rvert > 1, solution x \in (-\infty,-1) \cup (1,\infty)..

Q: Is x = 0 allowed in the domain?

No. x = 0 is not allowed because x^{-2} = 1/x^{2} is undefined at x = 0.

Q: What changes if the inequality is ≤ instead of <?

\frac{1}{x^{2}} \le 1 implies x^{2} \ge 1 , so |x| \ge 1 . The solution is x \in (-\infty,-1] \cup [1,\infty) , still excluding x = 0 .

Q: Why can we multiply both sides by x^{2} without flipping the inequality sign?

Why can we multiply both sides by x^{2} without flipping the inequality sign?

Q: Could you solve it by taking reciprocals instead?

Yes: 1/x^{2} < 1 \Leftrightarrow x^{2} > 1 by taking reciprocals of positive numbers, which leads to |x| > 1..

Q: What is the geometric interpretation?

Graph y = 1/x^{2}. The inequality 1/x^{2} < 1 asks where the curve lies below the horizontal line y = 1, which occurs when |x| > 1.

Answer

Interpret as \( |-(x^{-2})| \). Then
\( |-(x^{-2})| = \left|-\frac{1}{x^2}\right| = \frac{1}{x^2}\).
Solve \( \frac{1}{x^2} < 1\). Since \(x^2>0\), this is equivalent to \(x^2>1\), i.e. \(|x|>1\).

Final result: \(x\in(-\infty,-1)\cup(1,\infty)\).

Detailed Explanation

Problem

Solve the inequality

\[ \lvert -x^{-2}\rvert < 1 \]

Step-by-step solution

Write the expression inside the absolute value and simplify its sign.
\[ \lvert -x^{-2}\rvert = \lvert x^{-2}\rvert \]

Explanation: absolute value removes the sign, so the negative sign inside has no effect on the absolute value.
Note the domain.
\[ x^{-2} = \frac{1}{x^{2}} \text{ is undefined at } x=0. \]

Therefore we must have \(x \neq 0\). For all other real \(x\), \(x^{2}>0\) and hence \(\frac{1}{x^{2}}>0\).
Remove the absolute value using positivity of the inside (for \(x\neq 0\)).
\[ \lvert x^{-2}\rvert = x^{-2} = \frac{1}{x^{2}} \quad\text{for } x\neq 0. \]

Thus the inequality becomes

\[ \frac{1}{x^{2}} < 1. \]
Multiply both sides by \(x^{2}\).
Justification: for \(x\neq 0\) we have \(x^{2}>0\), so multiplication by \(x^{2}\) does not change the direction of the inequality.

Multiplying yields

\[ 1 < x^{2}. \]
Translate the inequality on \(x^{2}\) to an inequality on \(x\).
\[ x^{2} > 1 \iff \lvert x\rvert > 1. \]

That is, \(x<-1\) or \(x>1\).
Combine with the domain restriction.
The domain excluded \(x=0\), and the solution \(x<-1\) or \(x>1\) already excludes \(0\). Therefore the solution set is

\[ (-\infty,-1)\cup(1,\infty). \]

Final answer

\[ \{x\in\mathbb{R} : \lvert -x^{-2}\rvert < 1\} = (-\infty,-1)\cup(1,\infty). \]

See full solution

Ace your math homework with our AI tools - try it now!

Homework helper

Algebra FAQs

What does the notation \( -x^{-2} \) mean?

It means \( -x^{-2} = -\left(x^{-2}\right) = -1/x^{2} \); the exponent applies to \(x\) first, then the leading minus multiplies the result.

How do you simplify \( \lvert -x^{-2} \rvert \) ?

\( \lvert -x^{-2} \rvert = \lvert -1/x^{2} \rvert = 1/x^{2} \) for \( x \neq 0 \), since absolute value removes the negative sign..

How do you solve the inequality \( |-x^{-2}| < 1 \)?

Replace with \(1/x^{2} < 1\). Multiply by \(x^{2}>0\) ( \(x \neq 0\) ) to get \(1 < x^{2}\), so \(x^{2} > 1\). Hence \(\lvert x \rvert > 1\), solution \(x \in (-\infty,-1) \cup (1,\infty)\)..

Is \(x = 0\) allowed in the domain?

No. \(x = 0\) is not allowed because \(x^{-2} = 1/x^{2}\) is undefined at \(x = 0\).

What changes if the inequality is ≤ instead of

\( \frac{1}{x^{2}} \le 1 \) implies \( x^{2} \ge 1 \), so \( |x| \ge 1 \). The solution is \( x \in (-\infty,-1] \cup [1,\infty) \), still excluding \( x = 0 \).

Why can we multiply both sides by \(x^{2}\) without flipping the inequality sign?

Could you solve it by taking reciprocals instead?

Yes: \(1/x^{2} < 1 \Leftrightarrow x^{2} > 1\) by taking reciprocals of positive numbers, which leads to \(|x| > 1\)..

What is the geometric interpretation?

Graph \(y = 1/x^{2}\). The inequality \(1/x^{2} < 1\) asks where the curve lies below the horizontal line \(y = 1\), which occurs when \(|x| > 1\).

Solve complex math inequalities now
Try our finance and economics tools

252,312+ customers tried
Analytical, General, Biochemistry, etc.

Finance homework help Economics homework help Accounting homework help