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

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

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

extreme points

f (x) = (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)

y + 5 = \frac{1}{2} (x - 1)

f (x) = lo g_{2} (x)

\frac{x}{x ^{2} + x - 2}

s (t) = \frac{6 t}{sin ( t )}

f (x) = 0. 5^{x}

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