What is the extreme f(x)=x^3-6x^2+1 ?

The extreme f(x)=x^3-6x^2+1 is Maximum(0,1),Minimum(4,-31)

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extreme f(x)=x^3-6x^2+1

extreme points

f (x) = x^{3} - 6 x^{2} + 1

Maximum (0, 1), Minimum (4, - 31)

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} + 1 : 1

Maximum (0, 1)

Plug

x = 4

into

x^{3} - 6 x^{2} + 1 : - 31

Minimum (4, - 31)

Maximum (0, 1), Minimum (4, - 31)

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

f (x) = x^{4} - 2 x^{2} - 1

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

f (x, y) = 2 x^{4} + y^{2} - x^{2} - 2 y

f (x) = x ln (2 x)

What is the extreme f(x)=x^3-6x^2+1 ?
The extreme f(x)=x^3-6x^2+1 is Maximum(0,1),Minimum(4,-31)