What is the extreme f(x)=2x^3-4x^2-3 ?

The extreme f(x)=2x^3-4x^2-3 is Maximum(0,-3),Minimum(4/3 ,-145/27)

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extreme f(x)=2x^3-4x^2-3

extreme points

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

Maximum (0, - 3), Minimum (\frac{4}{3}, - \frac{145}{27})

Solution steps

f^{'} (x) = 6 x^{2} - 8 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{4}{3},

Increasing

: \frac{4}{3} < x < \infty

Plug

x = 0

into

2 x^{3} - 4 x^{2} - 3 : - 3

Maximum (0, - 3)

Plug

x = \frac{4}{3}

into

2 x^{3} - 4 x^{2} - 3 : - \frac{145}{27}

Minimum (\frac{4}{3}, - \frac{145}{27})

Maximum (0, - 3), Minimum (\frac{4}{3}, - \frac{145}{27})

f (x) = 2 - 2 cos (x)

\frac{1}{x ^{2}} + \frac{4}{y ^{2}} + 4 x y^{2}

f (x, y) = x^{2} + x y + y^{2} - 7 y + 16

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

What is the extreme f(x)=2x^3-4x^2-3 ?
The extreme f(x)=2x^3-4x^2-3 is Maximum(0,-3),Minimum(4/3 ,-145/27)