What is the extreme f(x)=(x+4)^3(3x-2)^2 ?

The extreme f(x)=(x+4)^3(3x-2)^2 is Saddle(-4,0),Maximum(-6/5 , 2151296/3125),Minimum(2/3 ,0)

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extreme f(x)=(x+4)^3(3x-2)^2

extreme points

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

Saddle (- 4, 0), Maximum (- \frac{6}{5}, \frac{2151296}{3125}), Minimum (\frac{2}{3}, 0)

Solution steps

f^{'} (x) = 3 (x + 4)^{2} (3 x - 2) (5 x + 6)

Find intervals:

Increasing

: - \infty < x < - 4,

Increasing

: - 4 < x < - \frac{6}{5},

Decreasing

: - \frac{6}{5} < x < \frac{2}{3},

Increasing

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

Plug

x = - 4

into

(x + 4)^{3} (3 x - 2)^{2} : 0

Saddle (- 4, 0)

Plug

x = - \frac{6}{5}

into

(x + 4)^{3} (3 x - 2)^{2} : \frac{2151296}{3125}

Maximum (- \frac{6}{5}, \frac{2151296}{3125})

Plug

x = \frac{2}{3}

into

(x + 4)^{3} (3 x - 2)^{2} : 0

Minimum (\frac{2}{3}, 0)

Saddle (- 4, 0), Maximum (- \frac{6}{5}, \frac{2151296}{3125}), Minimum (\frac{2}{3}, 0)

y = (x^{2} - 9)^{4}

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

f (x, y) = e^{y} - y e^{x}

f (x) = 3 x^{2} + 5 x - 4

f (x) = - \frac{1}{x ^{2}}

What is the extreme f(x)=(x+4)^3(3x-2)^2 ?
The extreme f(x)=(x+4)^3(3x-2)^2 is Saddle(-4,0),Maximum(-6/5 , 2151296/3125),Minimum(2/3 ,0)