What is the extreme f(x)=2x^3-12x^2+8 ?

The extreme f(x)=2x^3-12x^2+8 is Maximum(0,8),Minimum(4,-56)

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extreme f(x)=2x^3-12x^2+8

extreme points

f (x) = 2 x^{3} - 12 x^{2} + 8

Maximum (0, 8), Minimum (4, - 56)

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} + 8 : 8

Maximum (0, 8)

Plug

x = 4

into

2 x^{3} - 12 x^{2} + 8 : - 56

Minimum (4, - 56)

Maximum (0, 8), Minimum (4, - 56)

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

f (x, y) = x^{3} - y^{3} + 3 x^{2} + 3 y^{2} - 9 x

\frac{4 x}{x ^{2} - 36}

f (x, y) = x^{3} y^{2} + 5 y^{2} - x + 7

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

What is the extreme f(x)=2x^3-12x^2+8 ?
The extreme f(x)=2x^3-12x^2+8 is Maximum(0,8),Minimum(4,-56)