What is the extreme f(x)=x^2(5-4x)^2 ?

The extreme f(x)=x^2(5-4x)^2 is Minimum(0,0),Maximum(5/8 , 625/256),Minimum(5/4 ,0)

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extreme f(x)=x^2(5-4x)^2

extreme points

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

Minimum (0, 0), Maximum (\frac{5}{8}, \frac{625}{256}), Minimum (\frac{5}{4}, 0)

Solution steps

f^{'} (x) = 2 x (4 x - 5) (8 x - 5)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{5}{8},

Decreasing

: \frac{5}{8} < x < \frac{5}{4},

Increasing

: \frac{5}{4} < x < \infty

Plug

x = 0

into

x^{2} (5 - 4 x)^{2} : 0

Minimum (0, 0)

Plug

x = \frac{5}{8}

into

x^{2} (5 - 4 x)^{2} : \frac{625}{256}

Maximum (\frac{5}{8}, \frac{625}{256})

Plug

x = \frac{5}{4}

into

x^{2} (5 - 4 x)^{2} : 0

Minimum (\frac{5}{4}, 0)

Minimum (0, 0), Maximum (\frac{5}{8}, \frac{625}{256}), Minimum (\frac{5}{4}, 0)

p er p e n d i c u l a r y = 4 x - 7, (0, - 6)

r an g e f (x) = ∣ x ∣

d o main - lo g_{2} (x)

l in e (- 3, \frac{1}{32}), (3, 128)

in v erse f (x) = 3 \frac{1}{3} (x + 6) + 5

What is the extreme f(x)=x^2(5-4x)^2 ?
The extreme f(x)=x^2(5-4x)^2 is Minimum(0,0),Maximum(5/8 , 625/256),Minimum(5/4 ,0)