Find the value of x for which the function y = x\(^3\) - x has a minimum value.
\(-\sqrt{3}\)
\(-\frac{\sqrt{3}}{3}\)
\(\frac{\sqrt{3}}{3}\)
\(\sqrt{3}\)
Explanation
Video Explanation
No video available
Post your Contribution
Discussions (10)
dy/DX=0
y=x^3 -x
dy/DX=3x^2-1=0
3x^2-1=0
3x^2=1
x^2=1/3
x=√1/3=1/√3 * √3/3=√3/3
The first Derivatives is
y'= 3x²-1
at y'=0
3x²-1=0
3x²=1
x²=⅓
x=±√⅓. critical points
to obtain the Min.
y''= 6x
at x= +√⅓
y''= 6•√⅓ Min.
at x= -√⅓
y''= -6•√⅓ Max.
therefore, the value of x to make minimum point is √⅓ same as √3/3. MegaBodmas
First, find the derivative:
dy/dx = d(x^3 - x)/dx
= 3x^2 - 1
Next, set the derivative equal to 0 and solve for x:
3x^2 - 1 = 0
3x^2 = 1
x^2 = 1/3
x = ±√(1/3)
Now, to determine whether these values correspond to a minimum or maximum, we can use the second derivative test.
The second derivative is:
d^2y/dx^2 = d(3x^2 - 1)/dx
= 6x
Evaluating the second derivative at the critical points, we get:
at x = √(1/3):
d^2y/dx^2 = 6√(1/3) > 0 (minimum)
at x = -√(1/3):
d^2y/dx^2 = -6√(1/3) < 0 (maximum)
Therefore, the function y = x^3 - x has a minimum value at x = √(1/3).
Rationalizing the denominator, we get:
x = √(1/3)
= √(1/3) × √(3/3)
= √(3/9)
= √3/3
= 2/(3√3)
So, the value of x for which the function y = x^3 - x has a minimum value is indeed x = 2/(3√3).
Pls in my options I'm seeing all square root of 3 over 3 instead of square root of 3 over 3. This is confusing, it took my time before I could figure out which option to pick.
