What is the extreme f(x)=x^4-8x^3+16x^2 ?

The extreme f(x)=x^4-8x^3+16x^2 is Minimum(0,0),Maximum(2,16),Minimum(4,0)

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extreme f(x)=x^4-8x^3+16x^2

extreme points

f (x) = x^{4} - 8 x^{3} + 16 x^{2}

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

Solution steps

f^{'} (x) = 4 x^{3} - 24 x^{2} + 32 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

x^{4} - 8 x^{3} + 16 x^{2} : 0

Minimum (0, 0)

Plug

x = 2

into

x^{4} - 8 x^{3} + 16 x^{2} : 16

Maximum (2, 16)

Plug

x = 4

into

x^{4} - 8 x^{3} + 16 x^{2} : 0

Minimum (4, 0)

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

f (x) = 16 + 4 x - x^{2}, 0 \leq x \leq 5

f (x) = \frac{x ^{3}}{3} - x^{2} - 3 x - 1

f (x, y) = x^{4} + y^{4} - 16 x y

f (x) = x e^{\frac{- x ^{2}}{162}}

What is the extreme f(x)=x^4-8x^3+16x^2 ?
The extreme f(x)=x^4-8x^3+16x^2 is Minimum(0,0),Maximum(2,16),Minimum(4,0)