how do you solve x^3 ≤ 4x ?
回答 (9)
2009-09-05 6:43 pm
✔ 最佳答案
x^3 </= 4xx^3 - 4x </= 0
x ( x^2 - 4 ) </= 0.........(1)
...............................................................................
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 ].
........................................................................................
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 ].
................................................................................................
Combining the above two cases, the total solution set is
( -infinity, -2 ] U [ 0, 2 ]. .......................Ans.
.......................................................................................................
2009-09-05 6:38 pm
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, -â].
So, the solution set is [-2, -â] union [0,2].
Hope this helps.
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, -â].
So, the solution set is [-2, -â] union [0,2].
Hope this helps.
2009-09-05 6:53 pm
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 <= 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
2009-09-05 6:52 pm
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
ⴠx ⥠0,2 , x ⤠-2
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
ⴠx ⥠0,2 , x ⤠-2
2009-09-05 6:44 pm
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 -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.
2009-09-05 6:38 pm
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 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.
2009-09-05 6:37 pm
x^3-4x<0
x(x^2-4)<0
x(x+2)(x-2)<0
x<-2
2>x>0
x(x^2-4)<0
x(x+2)(x-2)<0
x<-2
2>x>0
2009-09-05 6:38 pm
divide each side by x:
x^2 </ 4
0</x</2 and x</-2
x^2 </ 4
0</x</2 and x</-2
2009-09-05 6:38 pm
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:
âx^2 = x
and
â4 = 2
so x ⤠2 is the solution
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:
âx^2 = x
and
â4 = 2
so x ⤠2 is the solution
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