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Find all \(x\) such that \(-4<{1\over{x}}<3\).

 Feb 14, 2022
 #1
avatar+361 
+1

ok..

i don't know if this is possible..

x can equal 1 to infinity and it works

 Feb 15, 2022
 #2
avatar+2437 
+1

Note that \(1 \over {1 \over3}\) is 3. When \(x\) gets bigger, the value of the expression gets smaller. 

 

Likewise, \(1 \over -{1 \over 4}\) is -4. As \(x\) gets smaller than \( -{1 \over 4}\), the value of the equation gets bigger, but never is greater than 0. 


So, the 2 ranges that work are \(\color{brown}\boxed{{x<-{1\over4} \space \text {and} \space{x>{1\over 3}}}}\)

BuilderBoi  Feb 15, 2022
edited by BuilderBoi  Feb 15, 2022
 #3
avatar+23198 
+1

Rewrite  -4  <  1/x  < 3  into its two parts:  -4  <  1/x   and   1/x  <  3

 

Divide the problem into two parts:

Part 1:  x < 0:

            (Remember that when you multiply both sides of an inequality by a negative number,

              you must change the sense of the inequality.

              Also remember that, for this part, x is negative.)

 

            -4 < 1/x   --->   -4x > 1   --->   x < -1/4

            1/x < 3   --->   1 > 3x   --->   3x < 1   --->   x < 1/3

 

            Since these two inequalities are combined by the word and, the solution to this

            part is the more restrictive of the two answers:  x < -1/4    

 

Part 2:  x > 0

 

             -4 < 1/x   --->   -4x < 1   --->   x > -1/4

             1/x < 3   --->   1 < 3x   --->   3x > 1   --->   x > 1/3

 

             The more restrictive of these two is:  x > 1/3

 

Since the answer consists of two parts, either Part 1 or Part 2, the final answer is the 

     region:  x < -1/4  or  x > 1/3

     which is:  (-inf, -1/4)  union  (1/3, inf)

 Feb 15, 2022

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