2

I can solve for $\frac 1 x \le 1$, but I cannot solve for $−5 < \frac 1 x$. Please help!

Jason
  • 21
  • 1
  • 5
    Distinguish two cases, a) $x > 0$, and b) $x < 0$. – Daniel Fischer Sep 08 '13 at 21:01
  • Note $$a<b \Rightarrow ac < bc \quad \text{iff}\ c > 0$$ $$a<b \Rightarrow ac > bc \quad \text{iff}\ c < 0$$ – AlexR Sep 08 '13 at 21:02
  • 2
    You could solve two seperate inequalities, that's -5<1/x and then 1/x<1 and then combine their solutions (with combine I mean intersection) Side note; did you graph the function? – imranfat Sep 08 '13 at 21:03

3 Answers3

2

My advice: draw the graph of $y = \frac1x$, and look at the places where the $y$-coordinate is between $-5$ and $1$.

This should give you an answer and help you visualize things. Then if you want to prove the answer algebraically, split into cases $x > 0$ and $x < 0$ as suggested in the comments. (What about the case $x = 0$?)

Caleb Stanford
  • 45,895
1

$$-5<\frac1x\le1\stackrel{\text{mult. by}\;x^2}\iff-5x^2<x\le x^2\iff\begin{cases}-5x^2<x\iff x(5x+1)>0\\{}\\\text{and}\\{}\\x\le x^2\iff x(x-1)\ge 0\end{cases}$$

Checking both parabolas we got above in the rightmost inequalities, we get:

$$\begin{align*}x(5x+1)>0&\iff x<-\frac15\;\;\vee\;\;x>0\\ x(x-1)\ge 0&\iff x\le 0\;\;\vee\;\;x\ge 1\end{align*}$$

The solution to the above "and" inequalities system is $\;x\le-\frac15\;\;\vee\;\;x\ge 1\;$

Observe carefully the first multiplication: this slick trick makes things simpler since then we don't have to distinguish between positive or negative multiplicatives: $\,x^2\,$ is always non-negative.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
1

Note that $\frac1x$ will never be zero for any real $x$. Hence, $$-5<\frac1x\le 1\qquad\Longleftrightarrow\qquad-5<\frac1x<0\quad\text{or}\quad0<\frac1x\le 1.$$ In both cases, we can simply multiply by $x,$ bearing in mind the sign of $x$ in each case.

Cameron Buie
  • 102,994