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

The extreme f(x)=x^3-6x^2+6 is Maximum(0,6),Minimum(4,-26)

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

extreme points

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

Maximum (0, 6), Minimum (4, - 26)

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} + 6 : 6

Maximum (0, 6)

Plug

x = 4

into

x^{3} - 6 x^{2} + 6 : - 26

Minimum (4, - 26)

Maximum (0, 6), Minimum (4, - 26)

f (x) = x^{2} + 4

f (x) = - 2 x^{3} + 33 x^{2} - 108 x + 1

f (x) = 5 x^{2} x + 1

f (x) = 3 cos^{2} (x), 0 \leq x \leq π

f (x) = - 8 x^{\frac{6}{5}} + 4 x^{\frac{2}{5}}

What is the extreme f(x)=x^3-6x^2+6 ?
The extreme f(x)=x^3-6x^2+6 is Maximum(0,6),Minimum(4,-26)