What is the extreme f(x)=2x^3-3x^2+7 ?

The extreme f(x)=2x^3-3x^2+7 is Maximum(0,7),Minimum(1,6)

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extreme f(x)=2x^3-3x^2+7

extreme points

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

Maximum (0, 7), Minimum (1, 6)

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} + 7 : 7

Maximum (0, 7)

Plug

x = 1

into

2 x^{3} - 3 x^{2} + 7 : 6

Minimum (1, 6)

Maximum (0, 7), Minimum (1, 6)

x y - x - 3 y

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

f (x) = x^{3} - 9 x^{2} + 1

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

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

What is the extreme f(x)=2x^3-3x^2+7 ?
The extreme f(x)=2x^3-3x^2+7 is Maximum(0,7),Minimum(1,6)