What is the extreme f(x)= 1/6 x^3-8x+19/3 ?

The extreme f(x)= 1/6 x^3-8x+19/3 is Maximum(-4, 83/3),Minimum(4,-15)

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extreme f(x)= 1/6 x^3-8x+19/3

extreme points

f (x) = \frac{1}{6} x^{3} - 8 x + \frac{19}{3}

Maximum (- 4, \frac{83}{3}), Minimum (4, - 15)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 4

into

\frac{1}{6} x^{3} - 8 x + \frac{19}{3} : \frac{83}{3}

Maximum (- 4, \frac{83}{3})

Plug

x = 4

into

\frac{1}{6} x^{3} - 8 x + \frac{19}{3} : - 15

Minimum (4, - 15)

Maximum (- 4, \frac{83}{3}), Minimum (4, - 15)

f (x) = x - ln (6 x)

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

- 7 x^{2} + 126 x - 560

20 - 0.06 x + 0.0002 x^{2}

f (x) = 2 x - 5 x^{2}

What is the extreme f(x)= 1/6 x^3-8x+19/3 ?
The extreme f(x)= 1/6 x^3-8x+19/3 is Maximum(-4, 83/3),Minimum(4,-15)