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

The extreme f(x)= 1/4 x^4+1/3 x^3-x^2+2 is Minimum(-2,-2/3),Maximum(0,2),Minimum(1, 19/12)

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

extreme points

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

Minimum (- 2, - \frac{2}{3}), Maximum (0, 2), Minimum (1, \frac{19}{12})

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} + 2 : - \frac{2}{3}

Minimum (- 2, - \frac{2}{3})

Plug

x = 0

into

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

Maximum (0, 2)

Plug

x = 1

into

\frac{1}{4} x^{4} + \frac{1}{3} x^{3} - x^{2} + 2 : \frac{19}{12}

Minimum (1, \frac{19}{12})

Minimum (- 2, - \frac{2}{3}), Maximum (0, 2), Minimum (1, \frac{19}{12})

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

f (x, y) = x^{2} + 2 y^{2} - 2 x - 4 y + 1

f (x, y) = 2 x^{2} + y^{2} + 8 x - 6 y - 2 x y + 12

f (x) = x^{3} - 3 x^{2} + 24 x

f (x) = \frac{2 x}{x - 1}

What is the extreme f(x)= 1/4 x^4+1/3 x^3-x^2+2 ?
The extreme f(x)= 1/4 x^4+1/3 x^3-x^2+2 is Minimum(-2,-2/3),Maximum(0,2),Minimum(1, 19/12)