What is the extreme f(x)=x^2(x-5)^2 ?

The extreme f(x)=x^2(x-5)^2 is Minimum(0,0),Maximum(5/2 , 625/16),Minimum(5,0)

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extreme f(x)=x^2(x-5)^2

extreme points

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

Minimum (0, 0), Maximum (\frac{5}{2}, \frac{625}{16}), Minimum (5, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

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

Decreasing

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

Increasing

: 5 < x < \infty

Plug

x = 0

into

x^{2} (x - 5)^{2} : 0

Minimum (0, 0)

Plug

x = \frac{5}{2}

into

x^{2} (x - 5)^{2} : \frac{625}{16}

Maximum (\frac{5}{2}, \frac{625}{16})

Plug

x = 5

into

x^{2} (x - 5)^{2} : 0

Minimum (5, 0)

Minimum (0, 0), Maximum (\frac{5}{2}, \frac{625}{16}), Minimum (5, 0)

What is the extreme f(x)=x^2(x-5)^2 ?
The extreme f(x)=x^2(x-5)^2 is Minimum(0,0),Maximum(5/2 , 625/16),Minimum(5,0)