What is the extreme f(x)=x^3-5x^2+3x+13 ?

The extreme f(x)=x^3-5x^2+3x+13 is Maximum(1/3 , 364/27),Minimum(3,4)

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extreme f(x)=x^3-5x^2+3x+13

extreme points

f (x) = x^{3} - 5 x^{2} + 3 x + 13

Maximum (\frac{1}{3}, \frac{364}{27}), Minimum (3, 4)

Solution steps

f^{'} (x) = 3 x^{2} - 10 x + 3

Find intervals:

Increasing

: - \infty < x < \frac{1}{3},

Decreasing

: \frac{1}{3} < x < 3,

Increasing

: 3 < x < \infty

Plug

x = \frac{1}{3}

into

x^{3} - 5 x^{2} + 3 x + 13 : \frac{364}{27}

Maximum (\frac{1}{3}, \frac{364}{27})

Plug

x = 3

into

x^{3} - 5 x^{2} + 3 x + 13 : 4

Minimum (3, 4)

Maximum (\frac{1}{3}, \frac{364}{27}), Minimum (3, 4)

f (x, y) = x + 4 y + \frac{2}{x y}

f (x, y) = x^{2} (x - 1) (2 x + 1) + y^{2}

f (x) = x^{3} - 3 x + 10

f (x, y) = 2 y^{3} - 3 x y + x^{2} - 1

- \frac{cos ( 2 x )}{2} - 2 sin (x)

What is the extreme f(x)=x^3-5x^2+3x+13 ?
The extreme f(x)=x^3-5x^2+3x+13 is Maximum(1/3 , 364/27),Minimum(3,4)