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

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

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

extreme points

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

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

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} + 4 : 4

Maximum (0, 4)

Plug

x = 2

into

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

Minimum (2, 0)

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

y = 2 x + 12

5 x + 2 y = 2

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

1 - \frac{2}{x + 3}

(1.4) (- 9.2)

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