# 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^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)**

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

x= (1/2)^(1/3).

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