What is the extreme f(x)=x^4-16x^3 ?

The extreme f(x)=x^4-16x^3 is Saddle(0,0),Minimum(12,-6912)

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extreme f(x)=x^4-16x^3

extreme points

f (x) = x^{4} - 16 x^{3}

Saddle (0, 0), Minimum (12, - 6912)

Solution steps

f^{'} (x) = 4 x^{3} - 48 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 12,

Increasing

: 12 < x < \infty

Plug

x = 0

into

x^{4} - 16 x^{3} : 0

Saddle (0, 0)

Plug

x = 12

into

x^{4} - 16 x^{3} : - 6912

Minimum (12, - 6912)

Saddle (0, 0), Minimum (12, - 6912)

f (x) = \frac{2 ^{2}}{2 ^{4} + 16}

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

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

f (x, y) = x y - 5 x + 15

f (x, y) = ln (4 x^{2} + 9 y^{2} + 36)

What is the extreme f(x)=x^4-16x^3 ?
The extreme f(x)=x^4-16x^3 is Saddle(0,0),Minimum(12,-6912)