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

The extreme f(x)=(-x^3)/3-2x^2+5x-2 is Minimum(-5,-106/3),Maximum(1, 2/3)

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

extreme points

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

Minimum (- 5, - \frac{106}{3}), Maximum (1, \frac{2}{3})

Solution steps

f^{'} (x) = - x^{2} - 4 x + 5

Find intervals:

Decreasing

: - \infty < x < - 5,

Increasing

: - 5 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 5

into

\frac{- x ^{3}}{3} - 2 x^{2} + 5 x - 2 : - \frac{106}{3}

Minimum (- 5, - \frac{106}{3})

Plug

x = 1

into

\frac{- x ^{3}}{3} - 2 x^{2} + 5 x - 2 : \frac{2}{3}

Maximum (1, \frac{2}{3})

Minimum (- 5, - \frac{106}{3}), Maximum (1, \frac{2}{3})

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

f (x, y) = \frac{1}{12} 144 - 16 x^{2} - 9 y^{2}

f (x) = 80 x^{2} - \frac{52}{3} x^{3} + x^{4}

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

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

What is the extreme f(x)=(-x^3)/3-2x^2+5x-2 ?
The extreme f(x)=(-x^3)/3-2x^2+5x-2 is Minimum(-5,-106/3),Maximum(1, 2/3)