What is the extreme y=(x^3(3x-8))/(12) ?

The extreme y=(x^3(3x-8))/(12) is Saddle(0,0),Minimum(2,-4/3)

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extreme y=(x^3(3x-8))/(12)

extreme points

y = \frac{x ^{3} ( 3 x - 8 )}{12}

Saddle (0, 0), Minimum (2, - \frac{4}{3})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

\frac{x ^{3} ( 3 x - 8 )}{12} : 0

Saddle (0, 0)

Plug

x = 2

into

\frac{x ^{3} ( 3 x - 8 )}{12} : - \frac{4}{3}

Minimum (2, - \frac{4}{3})

Saddle (0, 0), Minimum (2, - \frac{4}{3})

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

f (x) = e^{3 x} (7 - x)

\frac{( - x ^{2} - 8 x + 20 )}{( x ^{2} - 5 x ) ^{2}}

f (x) = 2 x^{2} + 1

f (x) = \frac{( 4 x ^{2} )}{x - 2}

What is the extreme y=(x^3(3x-8))/(12) ?
The extreme y=(x^3(3x-8))/(12) is Saddle(0,0),Minimum(2,-4/3)