What is the extreme f(x)=2x^3-12x^2+6 ?

The extreme f(x)=2x^3-12x^2+6 is Maximum(0,6),Minimum(4,-58)

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extreme f(x)=2x^3-12x^2+6

extreme points

f (x) = 2 x^{3} - 12 x^{2} + 6

Maximum (0, 6), Minimum (4, - 58)

Solution steps

f^{'} (x) = 6 x^{2} - 24 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

2 x^{3} - 12 x^{2} + 6 : 6

Maximum (0, 6)

Plug

x = 4

into

2 x^{3} - 12 x^{2} + 6 : - 58

Minimum (4, - 58)

Maximum (0, 6), Minimum (4, - 58)

f (x) = 2 x^{3} - 12 x^{2} + 4

f (x) = x^{3} - 6 x^{2} + 12 x + 4

f (x) = x^{\frac{2}{3}} - 1

f (x) = 48 x y - 32 x^{3} - 24 y^{2}

f (x) = x^{3} - 6 x^{2} + 12 x - 3

What is the extreme f(x)=2x^3-12x^2+6 ?
The extreme f(x)=2x^3-12x^2+6 is Maximum(0,6),Minimum(4,-58)