What is the extreme f(x)=x^4-4x^3+9 ?

The extreme f(x)=x^4-4x^3+9 is Saddle(0,9),Minimum(3,-18)

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extreme f(x)=x^4-4x^3+9

extreme points

f (x) = x^{4} - 4 x^{3} + 9

Saddle (0, 9), Minimum (3, - 18)

Solution steps

f^{'} (x) = 4 x^{3} - 12 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

x^{4} - 4 x^{3} + 9 : 9

Saddle (0, 9)

Plug

x = 3

into

x^{4} - 4 x^{3} + 9 : - 18

Minimum (3, - 18)

Saddle (0, 9), Minimum (3, - 18)

- x^{2} + 1

f (x) = 4 x^{3} + 7 x^{2} - 20 x + 9

f (x) = \frac{7}{x - 7}

2 x^{2} - 8

- x^{2} - 5

What is the extreme f(x)=x^4-4x^3+9 ?
The extreme f(x)=x^4-4x^3+9 is Saddle(0,9),Minimum(3,-18)