How to calculate the extreme points of the function
f(x)=x^3 + 3/x?
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To calculate the extreme points of the function, we have to differentiate the function, with respect to x.
We'll differentiate and we'll obtain:
f'(x) = (x^3)' + (3*x^-1)'
f'(x) = 3x^2 - 3*x^-2
f'(x) = 3x^2 - 3/x^2
f'(x) = (3x^4 - 3)/x^2
f'(x) = 3(x^4 - 1)/x^2
Now, we have to calculate the roots of the first derivative. If the first derivative is cancelling for a value of x, in that point the function has a stationary point.
We can see that the roots of the first derivative are;
(x^2 - 1)(x^2+1)=0
So, the extreme points of the function will be:
Extreme points: (1;4) and (-1;-4)
The extrme points of f(x) =x^3+3/x could be found by equating the derivative of f(x) to zero. Or solving for x for which f'(x) = 0.
f'(x) = (3x^2+3/x)'= 0. Or
3*2x+ (-3/x^2) = 0. Or
6x -3/x^2 = 0. Or
6x^3 -3 = 0. Or
x^3 = 1/2 . Or
To see whether tis is maxima or minima, we find whether f"(1/2) -negative or positive.
f''(x) = (6x-3/x^2) = 6+3*2/x^3 = 6+6/x^3 .
So f"(1/2) = 6+6/(1/2)^(1/3) which is positive.
So f(x) has a minimum at x= 1/2)^(1/3
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