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

The extreme f(x)=x^4+(4x^3)/3-4x^2 is Minimum(-2,-32/3),Maximum(0,0),Minimum(1,-5/3)

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

extreme points

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

Minimum (- 2, - \frac{32}{3}), Maximum (0, 0), Minimum (1, - \frac{5}{3})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

x^{4} + \frac{4 x ^{3}}{3} - 4 x^{2} : - \frac{32}{3}

Minimum (- 2, - \frac{32}{3})

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = 1

into

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

Minimum (1, - \frac{5}{3})

Minimum (- 2, - \frac{32}{3}), Maximum (0, 0), Minimum (1, - \frac{5}{3})

What is the extreme f(x)=x^4+(4x^3)/3-4x^2 ?
The extreme f(x)=x^4+(4x^3)/3-4x^2 is Minimum(-2,-32/3),Maximum(0,0),Minimum(1,-5/3)