What is the extreme f(x)= 1/3 x^3-x+9 ?

The extreme f(x)= 1/3 x^3-x+9 is Maximum(-1, 29/3),Minimum(1, 25/3)

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extreme f(x)= 1/3 x^3-x+9

extreme points

f (x) = \frac{1}{3} x^{3} - x + 9

Maximum (- 1, \frac{29}{3}), Minimum (1, \frac{25}{3})

Solution steps

f^{'} (x) = x^{2} - 1

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

\frac{1}{3} x^{3} - x + 9 : \frac{29}{3}

Maximum (- 1, \frac{29}{3})

Plug

x = 1

into

\frac{1}{3} x^{3} - x + 9 : \frac{25}{3}

Minimum (1, \frac{25}{3})

Maximum (- 1, \frac{29}{3}), Minimum (1, \frac{25}{3})

f (x) = - x^{3} - 15 x^{2}

f (x) = \frac{- x ^{2} - 6 x - 14}{x + 5}

f (x) = 17 + 4 x - x^{2}

y = 4 x^{2} + z^{2}

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

What is the extreme f(x)= 1/3 x^3-x+9 ?
The extreme f(x)= 1/3 x^3-x+9 is Maximum(-1, 29/3),Minimum(1, 25/3)