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

The extreme f(x)=x^3-3x^2+b is Maximum(0,b),Minimum(2,b-4)

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extreme f(x)=x^3-3x^2+b

extreme points

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

Maximum (0, b), Minimum (2, b - 4)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

x^{3} - 3 x^{2} + b : b

Maximum (0, b)

Plug

x = 2

into

x^{3} - 3 x^{2} + b : b - 4

Minimum (2, b - 4)

Maximum (0, b), Minimum (2, b - 4)

x^{3} + 6 x^{2} + 12 x + 2

f (x) = 27 x - x^{3}

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

f (x) = \frac{1}{x} + ln (x), e^{- 1} \leq x \leq e

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

What is the extreme f(x)=x^3-3x^2+b ?
The extreme f(x)=x^3-3x^2+b is Maximum(0,b),Minimum(2,b-4)