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

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

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

extreme points

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

Minimum (0, 0), Maximum (2, 16), Minimum (4, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = 2

into

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

Maximum (2, 16)

Plug

x = 4

into

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

Minimum (4, 0)

Minimum (0, 0), Maximum (2, 16), Minimum (4, 0)

p (x) = e^{x - 3} + e^{- x}

f (x, y) = x^{2} - 2 y + y

f (x, y) = 4 x^{2} + 8 y^{2}

f (x) = \frac{2 x + 1 - 18}{x}

f (x, y) = - 11 \cdot y^{2} + (x + 16)^{2} + 1

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