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

The extreme f(x)=(x^4)/4-x^3+3 is Saddle(0,3),Minimum(3,-15/4)

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

extreme

Saddle (0, 3), Minimum (3, - \frac{15}{4})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

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

Saddle (0, 3)

Plug

x = 3

into

\frac{x ^{4}}{4} - x^{3} + 3 : - \frac{15}{4}

Minimum (3, - \frac{15}{4})

Saddle (0, 3), Minimum (3, - \frac{15}{4})

What is the extreme f(x)=(x^4)/4-x^3+3 ?
The extreme f(x)=(x^4)/4-x^3+3 is Saddle(0,3),Minimum(3,-15/4)