What is the extreme f(x)=-2x^3+36x^2-192x ?

The extreme f(x)=-2x^3+36x^2-192x is Minimum(4,-320),Maximum(8,-256)

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extreme f(x)=-2x^3+36x^2-192x

extreme points

f (x) = - 2 x^{3} + 36 x^{2} - 192 x

Minimum (4, - 320), Maximum (8, - 256)

Solution steps

f^{'} (x) = - 6 x^{2} + 72 x - 192

Find intervals:

Decreasing

: - \infty < x < 4,

Increasing

: 4 < x < 8,

Decreasing

: 8 < x < \infty

Plug

x = 4

into

- 2 x^{3} + 36 x^{2} - 192 x : - 320

Minimum (4, - 320)

Plug

x = 8

into

- 2 x^{3} + 36 x^{2} - 192 x : - 256

Maximum (8, - 256)

Minimum (4, - 320), Maximum (8, - 256)

f (x) = 2 x^{3} - 6 x^{2} - 144 x + 7

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

f (x) = x (23 - 41 + 2 x) (\frac{41}{2} - x)

x^{3} - 3 x^{2} - 9 x + 2

x^{3} - 3 x^{2} - 9 x + 4

What is the extreme f(x)=-2x^3+36x^2-192x ?
The extreme f(x)=-2x^3+36x^2-192x is Minimum(4,-320),Maximum(8,-256)