What is the extreme f(x)=3x^4-6x^3+9 ?

The extreme f(x)=3x^4-6x^3+9 is Saddle(0,9),Minimum(3/2 , 63/16)

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extreme f(x)=3x^4-6x^3+9

extreme

Saddle (0, 9), Minimum (\frac{3}{2}, \frac{63}{16})

Solution steps

f^{'} (x) = 12 x^{3} - 18 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{3}{2},

Increasing

: \frac{3}{2} < x < \infty

Plug

x = 0

into

3 x^{4} - 6 x^{3} + 9 : 9

Saddle (0, 9)

Plug

x = \frac{3}{2}

into

3 x^{4} - 6 x^{3} + 9 : \frac{63}{16}

Minimum (\frac{3}{2}, \frac{63}{16})

Saddle (0, 9), Minimum (\frac{3}{2}, \frac{63}{16})

What is the extreme f(x)=3x^4-6x^3+9 ?
The extreme f(x)=3x^4-6x^3+9 is Saddle(0,9),Minimum(3/2 , 63/16)