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

The extreme f(x)=(x^3)/3-x^2-3x-1 is Maximum(-1, 2/3),Minimum(3,-10)

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

extreme

Maximum (- 1, \frac{2}{3}), Minimum (3, - 10)

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 - 1 : \frac{2}{3}

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

Plug

x = 3

into

\frac{x ^{3}}{3} - x^{2} - 3 x - 1 : - 10

Minimum (3, - 10)

Maximum (- 1, \frac{2}{3}), Minimum (3, - 10)

What is the extreme f(x)=(x^3)/3-x^2-3x-1 ?
The extreme f(x)=(x^3)/3-x^2-3x-1 is Maximum(-1, 2/3),Minimum(3,-10)