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

The extreme f(x)=(x-4)^2(x-2) is Maximum(8/3 , 32/27),Minimum(4,0)

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

extreme points

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

Maximum (\frac{8}{3}, \frac{32}{27}), Minimum (4, 0)

Solution steps

f^{'} (x) = (x - 4) (3 x - 8)

Find intervals:

Increasing

: - \infty < x < \frac{8}{3},

Decreasing

: \frac{8}{3} < x < 4,

Increasing

: 4 < x < \infty

Plug

x = \frac{8}{3}

into

(x - 4)^{2} (x - 2) : \frac{32}{27}

Maximum (\frac{8}{3}, \frac{32}{27})

Plug

x = 4

into

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

Minimum (4, 0)

Maximum (\frac{8}{3}, \frac{32}{27}), Minimum (4, 0)

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

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

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

x^{4} + 2 x^{3} + x^{2} + 2 x + 2

y = 3 x + 2 cos (x)

What is the extreme f(x)=(x-4)^2(x-2) ?
The extreme f(x)=(x-4)^2(x-2) is Maximum(8/3 , 32/27),Minimum(4,0)