What is the extreme f(x)=(x^3)/3-x^2-8x+3 ?

The extreme f(x)=(x^3)/3-x^2-8x+3 is Maximum(-2, 37/3),Minimum(4,-71/3)

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extreme f(x)=(x^3)/3-x^2-8x+3

extreme points

f (x) = \frac{x ^{3}}{3} - x^{2} - 8 x + 3

Maximum (- 2, \frac{37}{3}), Minimum (4, - \frac{71}{3})

Solution steps

f^{'} (x) = x^{2} - 2 x - 8

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 2

into

\frac{x ^{3}}{3} - x^{2} - 8 x + 3 : \frac{37}{3}

Maximum (- 2, \frac{37}{3})

Plug

x = 4

into

\frac{x ^{3}}{3} - x^{2} - 8 x + 3 : - \frac{71}{3}

Minimum (4, - \frac{71}{3})

Maximum (- 2, \frac{37}{3}), Minimum (4, - \frac{71}{3})

\frac{6 ∣ x ∣}{∣ x - 1 ∣}

f (x) = \frac{1}{3} x^{3} - \frac{3}{2} x^{2} - 18 x + \frac{1}{3}

\frac{2 x - 6}{x + 3}

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

f (x) = 4 x - 7 x^{\frac{4}{7}}

What is the extreme f(x)=(x^3)/3-x^2-8x+3 ?
The extreme f(x)=(x^3)/3-x^2-8x+3 is Maximum(-2, 37/3),Minimum(4,-71/3)