1/x ≤ 4/x^3
assuming x≠0
multiply both sides by x^3
For x>0 (=> x^3>0)
x²≤4 => x≤2
=> 0<x≤2
For x<0 (=> x^3<0), we need to reverse the direction of the inequality:
x²≥4 => x≤-2
=> -∞<x≤-2
So the final answer is:
0<x≤2 or -∞<x≤-2
In interval notations:
(-∞,-2]∪(0,2]
1/x is less than or equal to 4/x^3
3 answers
I got that same interval notation. Except I subtracted the 4/x^3 to the other side and ended up with (x+2)(x-2)/x^3 less than or equal to 0. Then I drew one of those. Line grAphs Nd put circles at -2,2, and an undefined at 0. Then figured out where the x's are negative. Would that be right?
ignoring the ≤ for the time being
1/x = 4/x^3
x^3 = 4x
x^3-4x=0
x(x^2-4) = 0
x(x+2)(x-2) = 0
so critical values are x = 0 and x = 2 , x = -2
4 regions to consider on the number line
1. x < -2 , let's say x = -4
Is 1/-4 ≤ 4/-64 ? YES
2. between -2 and 0 , let say x = -1
is 1/-1 ≤ 4/-1 , NO
3. between 0 and 2 , lets say x = 1.5
is .666 ≤ 1/1.185 ? , YES
4. x > 2 , lets say x = 4
is 1/4 ≤ 4/64 , NO
so x ≤ -2 or 0 < x ≤ 2
(notice x=0 is excluded, since we cannot divide by zero)
Wolfram illustrates the solution rather nicely
http://www.wolframalpha.com/input/?i=1%2Fx+%3D+4%2Fx%5E3
1/x = 4/x^3
x^3 = 4x
x^3-4x=0
x(x^2-4) = 0
x(x+2)(x-2) = 0
so critical values are x = 0 and x = 2 , x = -2
4 regions to consider on the number line
1. x < -2 , let's say x = -4
Is 1/-4 ≤ 4/-64 ? YES
2. between -2 and 0 , let say x = -1
is 1/-1 ≤ 4/-1 , NO
3. between 0 and 2 , lets say x = 1.5
is .666 ≤ 1/1.185 ? , YES
4. x > 2 , lets say x = 4
is 1/4 ≤ 4/64 , NO
so x ≤ -2 or 0 < x ≤ 2
(notice x=0 is excluded, since we cannot divide by zero)
Wolfram illustrates the solution rather nicely
http://www.wolframalpha.com/input/?i=1%2Fx+%3D+4%2Fx%5E3
1 answer