What is the extreme f(x)=(6-2x)(3-2x)x ?

The extreme f(x)=(6-2x)(3-2x)x is Maximum((3-sqrt(3))/2 ,3sqrt(3)),Minimum((3+sqrt(3))/2 ,-3sqrt(3))

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extreme f(x)=(6-2x)(3-2x)x

extreme points

f (x) = (6 - 2 x) (3 - 2 x) x

Maximum (\frac{3 - 3}{2}, 33), Minimum (\frac{3 + 3}{2}, - 33)

Solution steps

f^{'} (x) = 6 (2 x^{2} - 6 x + 3)

Find intervals:

Increasing

: - \infty < x < \frac{- 3 + 3}{2},

Decreasing

: \frac{- 3 + 3}{2} < x < \frac{3 + 3}{2},

Increasing

: \frac{3 + 3}{2} < x < \infty

Plug

x = \frac{3 - 3}{2}

into

(6 - 2 x) (3 - 2 x) x : 33

Maximum (\frac{3 - 3}{2}, 33)

Plug

x = \frac{3 + 3}{2}

into

(6 - 2 x) (3 - 2 x) x : - 33

Minimum (\frac{3 + 3}{2}, - 33)

Maximum (\frac{3 - 3}{2}, 33), Minimum (\frac{3 + 3}{2}, - 33)

\frac{x ^{3} + 30 x + 128}{x}, 10 \leq x \leq 20

e^{\frac{y ^{2}}{4} - x^{2} - 1}

f (x) = 6 x^{2} - 4 x + 1, 0 < x < 8

- x^{3} - 6 x^{2} + 3

f (x) = \frac{69120}{x} + \frac{69120}{y} + 5 x y

What is the extreme f(x)=(6-2x)(3-2x)x ?
The extreme f(x)=(6-2x)(3-2x)x is Maximum((3-sqrt(3))/2 ,3sqrt(3)),Minimum((3+sqrt(3))/2 ,-3sqrt(3))