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

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

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

extreme points

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

Maximum (0, 1), Minimum (2, - \frac{1}{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{x ^{3}}{3} - x^{2} + 1 : 1

Maximum (0, 1)

Plug

x = 2

into

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

Minimum (2, - \frac{1}{3})

Maximum (0, 1), Minimum (2, - \frac{1}{3})

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

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

f (x) = x + cos (x)

f (x) = - 2 x^{2} + 3 x + 7

f (x) = x^{2} - 5 x y + 6 y^{2} + 9 x - 5 y + 6

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