What is the extreme f(x)=(x^4)/4-(x^3)/3-3x^2 ?

The extreme f(x)=(x^4)/4-(x^3)/3-3x^2 is Minimum(-2,-16/3),Maximum(0,0),Minimum(3,-63/4)

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extreme f(x)=(x^4)/4-(x^3)/3-3x^2

extreme

Minimum (- 2, - \frac{16}{3}), Maximum (0, 0), Minimum (3, - \frac{63}{4})

Solution steps

f^{'} (x) = x^{3} - x^{2} - 6 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 2

into

\frac{x ^{4}}{4} - \frac{x ^{3}}{3} - 3 x^{2} : - \frac{16}{3}

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

Plug

x = 0

into

\frac{x ^{4}}{4} - \frac{x ^{3}}{3} - 3 x^{2} : 0

Maximum (0, 0)

Plug

x = 3

into

\frac{x ^{4}}{4} - \frac{x ^{3}}{3} - 3 x^{2} : - \frac{63}{4}

Minimum (3, - \frac{63}{4})

Minimum (- 2, - \frac{16}{3}), Maximum (0, 0), Minimum (3, - \frac{63}{4})

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