What is the extreme f(x)=x^4-12x^3+5 ?

The extreme f(x)=x^4-12x^3+5 is Saddle(0,5),Minimum(9,-2182)

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extreme f(x)=x^4-12x^3+5

extreme points

f (x) = x^{4} - 12 x^{3} + 5

Saddle (0, 5), Minimum (9, - 2182)

Solution steps

f^{'} (x) = 4 x^{3} - 36 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 9,

Increasing

: 9 < x < \infty

Plug

x = 0

into

x^{4} - 12 x^{3} + 5 : 5

Saddle (0, 5)

Plug

x = 9

into

x^{4} - 12 x^{3} + 5 : - 2182

Minimum (9, - 2182)

Saddle (0, 5), Minimum (9, - 2182)

x 64 - x^{2}

f (x, y) = x \cdot e^{\frac{y}{x}}

f (x, y) = x y + 4

f (x) = x + 8 - x

f (x) = x - \frac{16}{x}

What is the extreme f(x)=x^4-12x^3+5 ?
The extreme f(x)=x^4-12x^3+5 is Saddle(0,5),Minimum(9,-2182)