What is the extreme f(x)=2x^3-3x^2-6 ?

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

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extreme f(x)=2x^3-3x^2-6

extreme points

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

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

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} - 6 : - 6

Maximum (0, - 6)

Plug

x = 1

into

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

Minimum (1, - 7)

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

f (x) = \frac{( x ^{3} )}{( x - 1 ) ^{2}}

y = \frac{2 x ^{2}}{x - 6} [- 3]

f (x, y) = x - 2 y

3 x + 1 [3.6]

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