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

The extreme y=(x-3)^2(x+4)^2 is Minimum(-4,0),Maximum(-1/2 , 2401/16),Minimum(3,0)

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

extreme points

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

Minimum (- 4, 0), Maximum (- \frac{1}{2}, \frac{2401}{16}), Minimum (3, 0)

Solution steps

f^{'} (x) = 2 (x - 3) (x + 4) (2 x + 1)

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < - \frac{1}{2},

Decreasing

: - \frac{1}{2} < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 4

into

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

Minimum (- 4, 0)

Plug

x = - \frac{1}{2}

into

(x - 3)^{2} (x + 4)^{2} : \frac{2401}{16}

Maximum (- \frac{1}{2}, \frac{2401}{16})

Plug

x = 3

into

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

Minimum (3, 0)

Minimum (- 4, 0), Maximum (- \frac{1}{2}, \frac{2401}{16}), Minimum (3, 0)

f (x) = x - 4 - 4

6 x + 3 y = 2

f (x) = \frac{2 x - 8}{x ^{2} - 2 x - 3}

y = - 6 x + 8, (7, - 3)

What is the extreme y=(x-3)^2(x+4)^2 ?
The extreme y=(x-3)^2(x+4)^2 is Minimum(-4,0),Maximum(-1/2 , 2401/16),Minimum(3,0)