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

The extreme f(x)=-(x^2)/2+(x^3)/3 is Maximum(0,0),Minimum(1,-1/6)

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

extreme points

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

Maximum (0, 0), Minimum (1, - \frac{1}{6})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

- \frac{x ^{2}}{2} + \frac{x ^{3}}{3} : 0

Maximum (0, 0)

Plug

x = 1

into

- \frac{x ^{2}}{2} + \frac{x ^{3}}{3} : - \frac{1}{6}

Minimum (1, - \frac{1}{6})

Maximum (0, 0), Minimum (1, - \frac{1}{6})

What is the extreme f(x)=-(x^2)/2+(x^3)/3 ?
The extreme f(x)=-(x^2)/2+(x^3)/3 is Maximum(0,0),Minimum(1,-1/6)