What is the extreme f(x)=(x^3)/3-x^2-3x+11/3 ?

The extreme f(x)=(x^3)/3-x^2-3x+11/3 is Maximum(-1, 16/3),Minimum(3,-16/3)

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extreme f(x)=(x^3)/3-x^2-3x+11/3

extreme points

f (x) = \frac{x ^{3}}{3} - x^{2} - 3 x + \frac{11}{3}

Maximum (- 1, \frac{16}{3}), Minimum (3, - \frac{16}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 1

into

\frac{x ^{3}}{3} - x^{2} - 3 x + \frac{11}{3} : \frac{16}{3}

Maximum (- 1, \frac{16}{3})

Plug

x = 3

into

\frac{x ^{3}}{3} - x^{2} - 3 x + \frac{11}{3} : - \frac{16}{3}

Minimum (3, - \frac{16}{3})

Maximum (- 1, \frac{16}{3}), Minimum (3, - \frac{16}{3})

f (x, y) = 400 - 49 x^{2} - 81 v^{2}

f (x, y) = 3 x y - 2 x^{2} - 2 y^{2} + 28 x - 14 y - 3

f (x) = 2 x^{2} e^{- \frac{x}{5}}

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

f (x) = 9 - 2 x + 4 y - x^{2} - 2 y^{3}

What is the extreme f(x)=(x^3)/3-x^2-3x+11/3 ?
The extreme f(x)=(x^3)/3-x^2-3x+11/3 is Maximum(-1, 16/3),Minimum(3,-16/3)