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

The extreme f(x)=-x^2(x-4)(x+4) is Maximum(-2sqrt(2),64),Minimum(0,0),Maximum(2sqrt(2),64)

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

extreme points

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

Maximum (- 22, 64), Minimum (0, 0), Maximum (22, 64)

Solution steps

f^{'} (x) = - 4 x^{3} + 32 x

Find intervals:

Increasing

: - \infty < x < - 22,

Decreasing

: - 22 < x < 0,

Increasing

: 0 < x < 22,

Decreasing

: 22 < x < \infty

Plug

x = - 22

into

- x^{2} (x - 4) (x + 4) : 64

Maximum (- 22, 64)

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = 22

into

- x^{2} (x - 4) (x + 4) : 64

Maximum (22, 64)

Maximum (- 22, 64), Minimum (0, 0), Maximum (22, 64)

e^{9 x} + e^{- x}

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

x^{2} y - 6 y^{2} - 3 x^{2}

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

f (x) = x^{3} + x^{2} - 46 x + 80

What is the extreme f(x)=-x^2(x-4)(x+4) ?
The extreme f(x)=-x^2(x-4)(x+4) is Maximum(-2sqrt(2),64),Minimum(0,0),Maximum(2sqrt(2),64)