how do you solve x^3 ≤ 4x ?

x^3 </= 4x

x^3 - 4x </= 0

x ( x^2 - 4 ) </= 0.........(1)
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case(i)

If x < 0, then x^2 - 4 > 0, i.e., x^2 > 4, i.e., x < -2 or x > +2.

Hence, in this case, x is in ( -infinity, -2 ].
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case(ii)

If x > 0, then x^2 - 4 < 0, i.e., x^2 < 4, i.e., -2 < x < 2.

Hence, in this case, x is in [ 0, 2 ].
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Combining the above two cases, the total solution set is

( -infinity, -2 ] U [ 0, 2 ]. .......................Ans.
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x^3 ≤ 4x
x^3 - 4x ≤ 0
x(x^2 - 4) ≤ 0

This is possible only if x is positive and x^2 -4 is negative or if x is negative and if x^2 - 4 is positive. Let's analyze both cases.

Case 1: x ≥ 0 and x^2 - 4 ≤ 0.

Remember that x^2 is always positive. So, x^2 - 4 will be less than 0 only if |x| ≤ 2. As x is positive, this can be written as x ≥ 0, x ≤ 2 or
by the interval [0,2].

Case 2 : x ≤ 0 and x^2 - 4 ≥ 0.

Remember that x^2 is always positive. So, x^2 - 4 will be greater than 0 only if |x| ≥ 2. As x is negative, this can be written as x ≤ 0, x ≤ -2 or
by the interval [-2, -∞].

So, the solution set is [-2, -∞] union [0,2].

Hope this helps.

x^3 <= 4x

x^3 - 4x <= 0

x * (x^2 - 4) <= 0

In order for a product to be less than (or equal to) zero, one of the terms must be positive and one must be negative:

(a) x <= 0, (x^2 - 4) >= 0

x <= 0 and x^2 >= 4

x <= 0 and (x <= -2 or x>= 2)

x can't be less than 0 and greater than 2 at the same time. Also, if x <= 0 and x <= -2 at the same time, then x must be less than or equal to -2. Therefore, the above equation reduces to:

x <= -2

---------

(b) x >= 0, (x^2 - 4) <= 0

x >= 0 and x^2 <= 4

x >= 0 and (x >= -2 or x<= 2)

If x >= 0 and x >= -2 at the same time, then x must be greater than or equal to 0. Therefore, the above equation reduces to:

0 <= x <= 2

Combining the two answers:

x <= -2

or

0 <= x <= 2

x^3 ≤ 4x
x^3 - 4x ≤ 0
x(x^2 - 4) ≤ 0
x(x^2 - 2^2) ≤ 0
x(x + 2)(x - 2) ≤ 0

x ≥ 0

x + 2 ≤ 0
x ≤ -2

x - 2 ≥ 0
x ≥ 2

∴ x ≥ 0,2 , x ≤ -2

Then, x(x-2)(x+2)<or=0
x -2 0 2
________________________________________________
x+2 ------------------ 0 +++++++++++++++++++++++++++++++++
x ----------------------------------------------- 0 +++++++++++++++++++++
x-2 ----------------------------------------------------------------------- 0 ++++++++++
===================================================
Result -----------------0 ++++++++++ 0 -------------------- 0 +
x<-2 & 0<x<2
for x<= -2 & 0<=x<=2
God bless you.

x^3 ≤ 4x subtract 4x from both sides
x^3 - 4x ≤ 0 Factor an x out
x(x^2 -4) ≤ 0 then you have a difference of two squares
x(x-2)(x+2) ≤ 0 set each factor to 0
x ≤ 0 , x - 2 ≤ 0 , x + 2 ≤ 0 solve for x
x ≤ 0 , x ≤ 2 , x ≤ -2

sounds like it might be right.
so your answer is x ≤ 2

and dont go dividing both sides by x because you do not know what x is equal to because x may equal 0 and you cannot divide by zero.

x^3-4x<0
x(x^2-4)<0
x(x+2)(x-2)<0
x<-2
2>x>0

divide each side by x:

x^2 </ 4
0</x</2 and x</-2

From the equation you are shown x cubed is less than or equal to 4.
In these equations you want to be left with x on one side of the equation only so here you divide both sides by x:

x^3 / x = x^2
and
4x / x = 4

so now you know: x^2 ≤ 4

but you still want to be left with just x on one side so to get x you must square root both sides of the equation:

√x^2 = x
and
√4 = 2

so x ≤ 2 is the solution