What is the extreme y=x^4-4/3 x^3 ?

The extreme y=x^4-4/3 x^3 is Saddle(0,0),Minimum(1,-1/3)

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extreme y=x^4-4/3 x^3

extreme points

y = x^{4} - \frac{4}{3} x^{3}

Saddle (0, 0), Minimum (1, - \frac{1}{3})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

x^{4} - \frac{4}{3} x^{3} : 0

Saddle (0, 0)

Plug

x = 1

into

x^{4} - \frac{4}{3} x^{3} : - \frac{1}{3}

Minimum (1, - \frac{1}{3})

Saddle (0, 0), Minimum (1, - \frac{1}{3})

6 v + x

f (x) = 15 + 4 x - x^{2} [0.5]

f (x) = - 3 x + 5 ln (2 x), (1, 3)

x^{2} - y^{2} + 4 x + 5 y - 4 x y

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

What is the extreme y=x^4-4/3 x^3 ?
The extreme y=x^4-4/3 x^3 is Saddle(0,0),Minimum(1,-1/3)