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

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

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

extreme points

(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)

f (x) = 3 x^{2} - 12

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

f (x) = 14 x^{2} - 2 x^{3} 2 y^{2} + 4 x y

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

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

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