What is the extreme f(x)=(2x-2)^3(3x-1)^2 ?

The extreme f(x)=(2x-2)^3(3x-1)^2 is Maximum(1/3 ,0),Minimum(3/5 ,-1024/3125),Saddle(1,0)

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extreme f(x)=(2x-2)^3(3x-1)^2

extreme points

f (x) = (2 x - 2)^{3} (3 x - 1)^{2}

Maximum (\frac{1}{3}, 0), Minimum (\frac{3}{5}, - \frac{1024}{3125}), Saddle (1, 0)

Solution steps

f^{'} (x) = 24 (x - 1)^{2} (3 x - 1) (5 x - 3)

Find intervals:

Increasing

: - \infty < x < \frac{1}{3},

Decreasing

: \frac{1}{3} < x < \frac{3}{5},

Increasing

: \frac{3}{5} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = \frac{1}{3}

into

(2 x - 2)^{3} (3 x - 1)^{2} : 0

Maximum (\frac{1}{3}, 0)

Plug

x = \frac{3}{5}

into

(2 x - 2)^{3} (3 x - 1)^{2} : - \frac{1024}{3125}

Minimum (\frac{3}{5}, - \frac{1024}{3125})

Plug

x = 1

into

(2 x - 2)^{3} (3 x - 1)^{2} : 0

Saddle (1, 0)

Maximum (\frac{1}{3}, 0), Minimum (\frac{3}{5}, - \frac{1024}{3125}), Saddle (1, 0)

f (x) = x^{4} - 4 x^{3} - 8 x^{2}

f (x, y) = x^{3} ln (y^{2} + 1) - 2 y

f (x) = (\frac{x}{5})^{1} \cdot (\frac{y}{5})^{x - 1}

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

f (x) = x^{3} - 3 x^{2} + 10

What is the extreme f(x)=(2x-2)^3(3x-1)^2 ?
The extreme f(x)=(2x-2)^3(3x-1)^2 is Maximum(1/3 ,0),Minimum(3/5 ,-1024/3125),Saddle(1,0)