What is the extreme f(x)= 1/3 x^3-x^2+3 ?

The extreme f(x)= 1/3 x^3-x^2+3 is Maximum(0,3),Minimum(2, 5/3)

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extreme f(x)= 1/3 x^3-x^2+3

extreme points

f (x) = \frac{1}{3} x^{3} - x^{2} + 3

Maximum (0, 3), Minimum (2, \frac{5}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

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

Maximum (0, 3)

Plug

x = 2

into

\frac{1}{3} x^{3} - x^{2} + 3 : \frac{5}{3}

Minimum (2, \frac{5}{3})

Maximum (0, 3), Minimum (2, \frac{5}{3})

f (x) = \frac{1}{3} x^{3} - x^{2} + 1

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

f (x) = 3 cos (2 x) + 4

f (x) = 4 x^{3} - 12 x + 7

f (x) = 3 x + 6 x^{- 1}

What is the extreme f(x)= 1/3 x^3-x^2+3 ?
The extreme f(x)= 1/3 x^3-x^2+3 is Maximum(0,3),Minimum(2, 5/3)