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

The extreme f(x)=(x-3)^2(x+2) is Maximum(-1/3 , 500/27),Minimum(3,0)

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

extreme points

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

Maximum (- \frac{1}{3}, \frac{500}{27}), Minimum (3, 0)

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 3 < x < \infty

Plug

x = - \frac{1}{3}

into

(x - 3)^{2} (x + 2) : \frac{500}{27}

Maximum (- \frac{1}{3}, \frac{500}{27})

Plug

x = 3

into

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

Minimum (3, 0)

Maximum (- \frac{1}{3}, \frac{500}{27}), Minimum (3, 0)

f (x) = \frac{6 ( x - 7 ) ^{2}}{x + 9}

f (x) = - y - y^{3}

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

f (x) = \frac{( x ^{9} )}{e ^{2 x}}

f (x, y) = x^{2} - y^{2} + 21

What is the extreme f(x)=(x-3)^2(x+2) ?
The extreme f(x)=(x-3)^2(x+2) is Maximum(-1/3 , 500/27),Minimum(3,0)