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

The extreme y=(x-4)^2(x+2)^3 is Saddle(-2,0),Maximum(8/5 , 839808/3125),Minimum(4,0)

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

extreme points

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

Saddle (- 2, 0), Maximum (\frac{8}{5}, \frac{839808}{3125}), Minimum (4, 0)

Solution steps

f^{'} (x) = (x + 2)^{2} (x - 4) (5 x - 8)

Find intervals:

Increasing

: - \infty < x < - 2,

Increasing

: - 2 < x < \frac{8}{5},

Decreasing

: \frac{8}{5} < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 2

into

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

Saddle (- 2, 0)

Plug

x = \frac{8}{5}

into

(x - 4)^{2} (x + 2)^{3} : \frac{839808}{3125}

Maximum (\frac{8}{5}, \frac{839808}{3125})

Plug

x = 4

into

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

Minimum (4, 0)

Saddle (- 2, 0), Maximum (\frac{8}{5}, \frac{839808}{3125}), Minimum (4, 0)

f (x, y) = 2 x + y

f (x) = \frac{1}{3} x^{3} - x^{2} - 3 x + 3

f (x) = 5 x^{3} - 30 x^{2} + 45 x

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

f (x) = 3 x + \frac{1}{x}

What is the extreme y=(x-4)^2(x+2)^3 ?
The extreme y=(x-4)^2(x+2)^3 is Saddle(-2,0),Maximum(8/5 , 839808/3125),Minimum(4,0)