What is the extreme f(x)= 1/3 x^3+x^2-48x+20 ?

The extreme f(x)= 1/3 x^3+x^2-48x+20 is Maximum(-8, 892/3),Minimum(6,-160)

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extreme f(x)= 1/3 x^3+x^2-48x+20

extreme points

f (x) = \frac{1}{3} x^{3} + x^{2} - 48 x + 20

Maximum (- 8, \frac{892}{3}), Minimum (6, - 160)

Solution steps

f^{'} (x) = x^{2} + 2 x - 48

Find intervals:

Increasing

: - \infty < x < - 8,

Decreasing

: - 8 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = - 8

into

\frac{1}{3} x^{3} + x^{2} - 48 x + 20 : \frac{892}{3}

Maximum (- 8, \frac{892}{3})

Plug

x = 6

into

\frac{1}{3} x^{3} + x^{2} - 48 x + 20 : - 160

Minimum (6, - 160)

Maximum (- 8, \frac{892}{3}), Minimum (6, - 160)

f (x, y) = \frac{( x y )}{( x - y )}

f (x) = e^{x^{3} - 12 x + 5}

f (x, y) = \frac{π}{3} \cdot y \cdot (x^{2} + 2.8 \cdot x + 7.84)

y = x re

ln (x^{2} + 8)

What is the extreme f(x)= 1/3 x^3+x^2-48x+20 ?
The extreme f(x)= 1/3 x^3+x^2-48x+20 is Maximum(-8, 892/3),Minimum(6,-160)