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

The extreme f(x)=x^3-6x^2+2 is Maximum(0,2),Minimum(4,-30)

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

extreme points

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

Maximum (0, 2), Minimum (4, - 30)

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} + 2 : 2

Maximum (0, 2)

Plug

x = 4

into

x^{3} - 6 x^{2} + 2 : - 30

Minimum (4, - 30)

Maximum (0, 2), Minimum (4, - 30)

f (l) = l

f (x) = \frac{7}{x - 4}

f (x) = \frac{1}{x + 13}

f (x) = \frac{- x - 3}{2 x - 6}

(\frac{1}{2}, 4), (\frac{3}{2}, - 1)

What is the extreme f(x)=x^3-6x^2+2 ?
The extreme f(x)=x^3-6x^2+2 is Maximum(0,2),Minimum(4,-30)