What is the extreme f(x)=(-2)/3 x^3+10x^2-48x-1 ?

The extreme f(x)=(-2)/3 x^3+10x^2-48x-1 is Minimum(4,-227/3),Maximum(6,-73)

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extreme f(x)=(-2)/3 x^3+10x^2-48x-1

extreme points

f (x) = \frac{- 2}{3} x^{3} + 10 x^{2} - 48 x - 1

Minimum (4, - \frac{227}{3}), Maximum (6, - 73)

Solution steps

f^{'} (x) = - 2 x^{2} + 20 x - 48

Find intervals:

Decreasing

: - \infty < x < 4,

Increasing

: 4 < x < 6,

Decreasing

: 6 < x < \infty

Plug

x = 4

into

\frac{- 2}{3} x^{3} + 10 x^{2} - 48 x - 1 : - \frac{227}{3}

Minimum (4, - \frac{227}{3})

Plug

x = 6

into

\frac{- 2}{3} x^{3} + 10 x^{2} - 48 x - 1 : - 73

Maximum (6, - 73)

Minimum (4, - \frac{227}{3}), Maximum (6, - 73)

x^{\frac{2}{3}} (x^{3} - 44)

2 x^{2} + 16 x - 4

\frac{x ^{5}}{20} - 3 x^{3}

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

f (x) = 8 x y - 2 x^{4} - 2 y^{4}

What is the extreme f(x)=(-2)/3 x^3+10x^2-48x-1 ?
The extreme f(x)=(-2)/3 x^3+10x^2-48x-1 is Minimum(4,-227/3),Maximum(6,-73)