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

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

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

extreme points

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

Minimum (0, 0), Maximum (2, 16), Minimum (3, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

x^{4} (x - 3)^{2} : 0

Minimum (0, 0)

Plug

x = 2

into

x^{4} (x - 3)^{2} : 16

Maximum (2, 16)

Plug

x = 3

into

x^{4} (x - 3)^{2} : 0

Minimum (3, 0)

Minimum (0, 0), Maximum (2, 16), Minimum (3, 0)

f (x, y) = 40 x^{\frac{3}{4}} y^{\frac{1}{4}}

y = 9 e^{- 2 t} - 7 e^{- 3 t}

18 x - 8

x^{6} - x^{4} + 4 x^{3} - 2 x

f (x) = (x^{2} - 4)^{4} (x^{2} + 1)^{5}

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