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

The extreme f(x)=x^2(x-2)^3 is Maximum(0,0),Minimum(4/5 ,-3456/3125),Saddle(2,0)

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

extreme points

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

Maximum (0, 0), Minimum (\frac{4}{5}, - \frac{3456}{3125}), Saddle (2, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{4}{5},

Increasing

: \frac{4}{5} < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = \frac{4}{5}

into

x^{2} (x - 2)^{3} : - \frac{3456}{3125}

Minimum (\frac{4}{5}, - \frac{3456}{3125})

Plug

x = 2

into

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

Saddle (2, 0)

Maximum (0, 0), Minimum (\frac{4}{5}, - \frac{3456}{3125}), Saddle (2, 0)

f (x) = \frac{6 x ^{5}}{5} - 15 x^{3} - 81 x + 2

- 0.1 t^{2} 0.8 t + 98.8

f (x, y) = x^{3} + y^{3} - 108 x - 27 y - 6

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

f (x) = cos (π x), 0 \leq x \leq \frac{1}{6}

What is the extreme f(x)=x^2(x-2)^3 ?
The extreme f(x)=x^2(x-2)^3 is Maximum(0,0),Minimum(4/5 ,-3456/3125),Saddle(2,0)