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) = 2 x^{3} - 4 x^{2} + 2 x + 1

4 - 3 x - x^{3}

f (x) = 4 - x^{3}

- 2 x^{2} + 4 x + 8

f (x y) = x^{2} + 2 y^{2} - x y + 14 y

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)