How to calculate the extreme points of the function f(x)=x^3 + 3/x?

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^4-1=0

(x^2 - 1)(x^2+1)=0

(x-1)(x+1)(x^2+1)=0

x-1=0

x1=1

x+1=0

x2=-1

x^2+1>0

So, the extreme points of the function will be:

f(1)=1+3=4

f(-1)=-1-3=-4

Extreme points: (1;4) and (-1;-4)