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(1,3)

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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, 3)

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 : 3

Minimum (1, 3)

Maximum (0, 4), Minimum (1, 3)

2 x^{3} + 6 x y^{2} - 3 y^{3} - 150 x

f (x) = x^{3} - 2 x^{2} - 15 x + 10, - 2 \leq x \leq 0

f (x, y) = 2 x^{4} - 5 x^{2} y^{3} - 3 x y^{4}

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

f (x) = x^{3} - 9 x^{2} + 4

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(1,3)