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

The extreme f(x)=x^3-6x^2 is Maximum(0,0),Minimum(4,-32)

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

extreme points

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

Maximum (0, 0), Minimum (4, - 32)

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} : 0

Maximum (0, 0)

Plug

x = 4

into

x^{3} - 6 x^{2} : - 32

Minimum (4, - 32)

Maximum (0, 0), Minimum (4, - 32)

y = - \frac{4}{5} x + \frac{9}{5}, (- 5, - 5)

f (x) = x ln (x)

r (n) = \frac{1}{n - 1}

e^{x^{2} - 2 x - 3}

f (x) = \frac{x - 13}{2}

What is the extreme f(x)=x^3-6x^2 ?
The extreme f(x)=x^3-6x^2 is Maximum(0,0),Minimum(4,-32)