What is the extreme f(x)=x^4-8/3 x^3+10 ?

The extreme f(x)=x^4-8/3 x^3+10 is Saddle(0,10),Minimum(2, 14/3)

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extreme f(x)=x^4-8/3 x^3+10

extreme points

f (x) = x^{4} - \frac{8}{3} x^{3} + 10

Saddle (0, 10), Minimum (2, \frac{14}{3})

Solution steps

f^{'} (x) = 4 x^{3} - 8 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

x^{4} - \frac{8}{3} x^{3} + 10 : 10

Saddle (0, 10)

Plug

x = 2

into

x^{4} - \frac{8}{3} x^{3} + 10 : \frac{14}{3}

Minimum (2, \frac{14}{3})

Saddle (0, 10), Minimum (2, \frac{14}{3})

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

x^{2} - y^{2}

f (x, y) = ln (x + y + 1)

K (r, s) = 4 r - 7 s

f (x) = \frac{x}{81 + x ^{2}}

What is the extreme f(x)=x^4-8/3 x^3+10 ?
The extreme f(x)=x^4-8/3 x^3+10 is Saddle(0,10),Minimum(2, 14/3)