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

The extreme f(x)=(x^4)/4-3x^3+4x^2 is Minimum(0,0),Maximum(1, 5/4),Minimum(8,-256)

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

extreme points

f (x) = \frac{x ^{4}}{4} - 3 x^{3} + 4 x^{2}

Minimum (0, 0), Maximum (1, \frac{5}{4}), Minimum (8, - 256)

Solution steps

f^{'} (x) = x^{3} - 9 x^{2} + 8 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = 1

into

\frac{x ^{4}}{4} - 3 x^{3} + 4 x^{2} : \frac{5}{4}

Maximum (1, \frac{5}{4})

Plug

x = 8

into

\frac{x ^{4}}{4} - 3 x^{3} + 4 x^{2} : - 256

Minimum (8, - 256)

Minimum (0, 0), Maximum (1, \frac{5}{4}), Minimum (8, - 256)

f (x) = 5 x - ln (x)

f (x) = x^{4} - 12 x^{3} + 16 x^{2} + 5

f (x) = 4 x^{3} - 2 y^{2} - 12 x + 8 y

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

f (x) = - x^{3} + 3 x^{2} + 24 x - 4

What is the extreme f(x)=(x^4)/4-3x^3+4x^2 ?
The extreme f(x)=(x^4)/4-3x^3+4x^2 is Minimum(0,0),Maximum(1, 5/4),Minimum(8,-256)