What is the extreme f(x)=(x^4)/4+5/3 x^3-5x^2-50x+1 ?

The extreme f(x)=(x^4)/4+5/3 x^3-5x^2-50x+1 is

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extreme f(x)=(x^4)/4+5/3 x^3-5x^2-50x+1

extreme points

f (x) = \frac{x ^{4}}{4} + \frac{5}{3} x^{3} - 5 x^{2} - 50 x + 1

Minimum (- 5, \frac{887}{12}), Maximum (- 10, - \frac{50 10}{3} + 5010 - 24), Minimum (10, \frac{50 10}{3} - 24 - 5010)

Solution steps

f^{'} (x) = x^{3} + 5 x^{2} - 10 x - 50

Find intervals:

Decreasing

: - \infty < x < - 5,

Increasing

: - 5 < x < - 10,

Decreasing

: - 10 < x < 10,

Increasing

: 10 < x < \infty

Plug

x = - 5

into

\frac{x ^{4}}{4} + \frac{5}{3} x^{3} - 5 x^{2} - 50 x + 1 : \frac{887}{12}

Minimum (- 5, \frac{887}{12})

Plug

x = - 10

into

\frac{x ^{4}}{4} + \frac{5}{3} x^{3} - 5 x^{2} - 50 x + 1 : - \frac{50 10}{3} + 5010 - 24

Maximum (- 10, - \frac{50 10}{3} + 5010 - 24)

Plug

x = 10

into

\frac{x ^{4}}{4} + \frac{5}{3} x^{3} - 5 x^{2} - 50 x + 1 : \frac{50 10}{3} - 24 - 5010

Minimum (10, \frac{50 10}{3} - 24 - 5010)

Minimum (- 5, \frac{887}{12}), Maximum (- 10, - \frac{50 10}{3} + 5010 - 24), Minimum (10, \frac{50 10}{3} - 24 - 5010)

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

f (x) = 2 x^{2} - 8 x + 2 (0.5)

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

f (x, y) = x^{3} - 12 x + y^{2} + 6 y + 14

f (x) = \frac{1}{5} x^{5} - \frac{2}{3} x^{3} + x

What is the extreme f(x)=(x^4)/4+5/3 x^3-5x^2-50x+1 ?
The extreme f(x)=(x^4)/4+5/3 x^3-5x^2-50x+1 is