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

The extreme f(x)=2x^3-3x^2-4 is Maximum(0,-4),Minimum(1,-5)

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

extreme points

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

Maximum (0, - 4), Minimum (1, - 5)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

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

Maximum (0, - 4)

Plug

x = 1

into

2 x^{3} - 3 x^{2} - 4 : - 5

Minimum (1, - 5)

Maximum (0, - 4), Minimum (1, - 5)

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

f (x) = x^{3} - 2 x^{2} - 15 x + 10 [- 2]

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

y = \frac{1}{2} x^{2} + (\frac{14 - 3 x}{4})^{2}

93000 x - 36000

What is the extreme f(x)=2x^3-3x^2-4 ?
The extreme f(x)=2x^3-3x^2-4 is Maximum(0,-4),Minimum(1,-5)