What is the extreme y=(x-2)^2(x-1) ?

The extreme y=(x-2)^2(x-1) is Maximum(4/3 , 4/27),Minimum(2,0)

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extreme y=(x-2)^2(x-1)

extreme points

y = (x - 2)^{2} (x - 1)

Maximum (\frac{4}{3}, \frac{4}{27}), Minimum (2, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < \frac{4}{3},

Decreasing

: \frac{4}{3} < x < 2,

Increasing

: 2 < x < \infty

Plug

x = \frac{4}{3}

into

(x - 2)^{2} (x - 1) : \frac{4}{27}

Maximum (\frac{4}{3}, \frac{4}{27})

Plug

x = 2

into

(x - 2)^{2} (x - 1) : 0

Minimum (2, 0)

Maximum (\frac{4}{3}, \frac{4}{27}), Minimum (2, 0)

f (x) = \frac{x ^{3}}{3} - x^{2} - 3 x + \frac{11}{3}

f (x, y) = 400 - 49 x^{2} - 81 v^{2}

f (x, y) = 3 x y - 2 x^{2} - 2 y^{2} + 28 x - 14 y - 3

f (x) = 2 x^{2} e^{- \frac{x}{5}}

f (x, y) = 3 x y + x^{3} - y^{3} + 1

What is the extreme y=(x-2)^2(x-1) ?
The extreme y=(x-2)^2(x-1) is Maximum(4/3 , 4/27),Minimum(2,0)