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Solve the inequality x^3 + 4x > 5x.

 Jul 10, 2021
 #1
avatar+357 
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Solve the inequality x^3 + 4x > 5x.

 

x^3-x>0

x(x+2)(x-1)>0

-1x

 

x>1

 

Yay!

 Jul 10, 2021
 #2
avatar+373 
+4

Hey there, Guest!

 

Let's find the critical points of the inequality:

 

\(x^3+4x=5x\)

 

\(x^3+4x−5x=5x−5x\) (Subtract 5x from both sides)

 

\(x^3−x=0 \)

 

\(x(x+1)(x−1)=0\) (Factor left side of the equation)

 

\(x=0\) or \(x=-1\) or \(x=1\)

 

Check intervals in between critical points. (Test values in the intervals to see if they work.)

 

\(x<−1\) (Doesn't work in original inequality)

 

\(−1  (Works in original inequality)

 

\(0  (Doesn't work in original inequality)

 

\(x>1\) (Works in original inequality)

 

Therefore the answers to your equation are: 

 

−1  or  x>1

 

Hope this helped! :)

( ゚д゚)つ Bye

 Jul 10, 2021

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