What is the extreme f(x)=x^3-6x^2+13 ?

The extreme f(x)=x^3-6x^2+13 is Maximum(0,13),Minimum(4,-19)

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extreme f(x)=x^3-6x^2+13

extreme points

f (x) = x^{3} - 6 x^{2} + 13

Maximum (0, 13), Minimum (4, - 19)

Solution steps

f^{'} (x) = 3 x^{2} - 12 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

x^{3} - 6 x^{2} + 13 : 13

Maximum (0, 13)

Plug

x = 4

into

x^{3} - 6 x^{2} + 13 : - 19

Minimum (4, - 19)

Maximum (0, 13), Minimum (4, - 19)

f (x) = 7 x - 2 y = - 2

g (x) = 4 x + 48

x^{3} - 6 x^{2} + 9 x

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

(x + 2)^{2} + 1

What is the extreme f(x)=x^3-6x^2+13 ?
The extreme f(x)=x^3-6x^2+13 is Maximum(0,13),Minimum(4,-19)