What is the extreme y=(x-9)*sqrt(2x) ?

The extreme y=(x-9)*sqrt(2x) is Maximum(0,0),Minimum(3,-6sqrt(6))

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extreme y=(x-9)*sqrt(2x)

extreme points

y = (x - 9) \cdot 2 x

Maximum (0, 0), Minimum (3, - 66)

Solution steps

f^{'} (x) = \frac{3 x - 9}{2 x}

Find intervals:

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

(x - 9) \cdot 2 x : 0

Maximum (0, 0)

Plug

x = 3

into

(x - 9) \cdot 2 x : - 66

Minimum (3, - 66)

Maximum (0, 0), Minimum (3, - 66)

f (x) = x y^{2} - y^{2} + \frac{1}{2} x^{2} - 5 x + 4

0.001 x^{2} + 7 + \frac{128}{x}

f (t) = \frac{1}{9} [\frac{t ^{3}}{3} + \frac{3 t ^{2}}{2} - 28 t]

F (X, Y) = X \cdot Y

What is the extreme y=(x-9)*sqrt(2x) ?
The extreme y=(x-9)*sqrt(2x) is Maximum(0,0),Minimum(3,-6sqrt(6))