Solve the inequality: `x^3-x<=0`

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jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

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Let f(x)=(x^3-x) =x(x^2-1) = x(x-1)(x+1)

(x^3-x) <=0

x(x-1)(x+1)<=0

At x=0,1,-1; f(x)=0

When x<-1(put x=-2); f(x) <0

When -1<x<0 (put x=-0.5) ; f(x)>0

When 0<x<1 (put x=0.5) ; f(x)<0

When 1<x (put x=2) ; f(x)>0

So the equation satisfies at x<-1 and 0<x<1

So the answer is x ε(-∞,-1] U [0,1]

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The inequality `x^3 - x <= 0` has to be solved.

`x^3 - x <= 0`

=> `x(x^2 - 1) <= 0`

This is true when either `x <= 0` and `x^2 - 1 >= 0` or when `x >=0` and `x^2 - 1 <= 0`

=> `x <= 0` and `x^2 >= 1` or `x >= 0` and `x^2 <= 1`

=> `x <= -1` or `0 <= x <= 1`

The inequality is satisfied for x lying in `{-oo, -1]U[0, 1]`

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