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

The extreme f(x)= 1/4 x^4+1/3 x^3-x^2+2/3 is Minimum(-2,-2),Maximum(0, 2/3),Minimum(1, 1/4)

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

extreme points

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

Minimum (- 2, - 2), Maximum (0, \frac{2}{3}), Minimum (1, \frac{1}{4})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

\frac{1}{4} x^{4} + \frac{1}{3} x^{3} - x^{2} + \frac{2}{3} : - 2

Minimum (- 2, - 2)

Plug

x = 0

into

\frac{1}{4} x^{4} + \frac{1}{3} x^{3} - x^{2} + \frac{2}{3} : \frac{2}{3}

Maximum (0, \frac{2}{3})

Plug

x = 1

into

\frac{1}{4} x^{4} + \frac{1}{3} x^{3} - x^{2} + \frac{2}{3} : \frac{1}{4}

Minimum (1, \frac{1}{4})

Minimum (- 2, - 2), Maximum (0, \frac{2}{3}), Minimum (1, \frac{1}{4})

f (x, y) = - 4 \cdot \frac{x ^{2} + 6 x + 8}{y ^{2} + 6 y + 6}

f (x) = x + \frac{108}{x ^{2}}

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

g (x, y) = x y^{\frac{1}{2}}

f (x) = x^{3} - 6 x^{2} + 12 x - 5

What is the extreme f(x)= 1/4 x^4+1/3 x^3-x^2+2/3 ?
The extreme f(x)= 1/4 x^4+1/3 x^3-x^2+2/3 is Minimum(-2,-2),Maximum(0, 2/3),Minimum(1, 1/4)