What is the extreme f(x)=x^3-6x^2+4 ?

The extreme f(x)=x^3-6x^2+4 is Maximum(0,4),Minimum(4,-28)

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extreme f(x)=x^3-6x^2+4

extreme points

f (x) = x^{3} - 6 x^{2} + 4

Maximum (0, 4), Minimum (4, - 28)

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} + 4 : 4

Maximum (0, 4)

Plug

x = 4

into

x^{3} - 6 x^{2} + 4 : - 28

Minimum (4, - 28)

Maximum (0, 4), Minimum (4, - 28)

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

f (x) = x^{2} - 2 x - 8

f (x) = \frac{- 8}{x ^{2} - 4}

f (x) = \frac{x ^{2} - x - 2}{x ^{2} - 6 x + 9}

f (x, y) = x y + \frac{8}{x} + \frac{8}{y}

What is the extreme f(x)=x^3-6x^2+4 ?
The extreme f(x)=x^3-6x^2+4 is Maximum(0,4),Minimum(4,-28)