What is the extreme y=x^3-30x^2+6000 ?

The extreme y=x^3-30x^2+6000 is Maximum(0,6000),Minimum(20,2000)

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extreme y=x^3-30x^2+6000

extreme points

y = x^{3} - 30 x^{2} + 6000

Maximum (0, 6000), Minimum (20, 2000)

Solution steps

f^{'} (x) = 3 x^{2} - 60 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 20,

Increasing

: 20 < x < \infty

Plug

x = 0

into

x^{3} - 30 x^{2} + 6000 : 6000

Maximum (0, 6000)

Plug

x = 20

into

x^{3} - 30 x^{2} + 6000 : 2000

Minimum (20, 2000)

Maximum (0, 6000), Minimum (20, 2000)

f (x) = x^{2} + x y + y^{2} + y

f (x) - x^{4} - 8 x^{2} + 1

f (x) = x^{3} - 12 x + 17

f (x) = x^{3} + 2 x^{2} - 20 x

e^{4 x} + e^{- x}

What is the extreme y=x^3-30x^2+6000 ?
The extreme y=x^3-30x^2+6000 is Maximum(0,6000),Minimum(20,2000)