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

The extreme f(x)=x^4+2x^3-2x^2+2 is Minimum(-2,-6),Maximum(0,2),Minimum(1/2 , 29/16)

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

extreme points

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

Minimum (- 2, - 6), Maximum (0, 2), Minimum (\frac{1}{2}, \frac{29}{16})

Solution steps

f^{'} (x) = 4 x^{3} + 6 x^{2} - 4 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = - 2

into

x^{4} + 2 x^{3} - 2 x^{2} + 2 : - 6

Minimum (- 2, - 6)

Plug

x = 0

into

x^{4} + 2 x^{3} - 2 x^{2} + 2 : 2

Maximum (0, 2)

Plug

x = \frac{1}{2}

into

x^{4} + 2 x^{3} - 2 x^{2} + 2 : \frac{29}{16}

Minimum (\frac{1}{2}, \frac{29}{16})

Minimum (- 2, - 6), Maximum (0, 2), Minimum (\frac{1}{2}, \frac{29}{16})

f (x) = 5 - x^{2}

x^{4} - 5 x^{3}

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 3, - 1 \leq x \leq 10

f (x) = \frac{8}{3 x ^{2} + 4}

f (x) = 2 sin (x)

What is the extreme f(x)=x^4+2x^3-2x^2+2 ?
The extreme f(x)=x^4+2x^3-2x^2+2 is Minimum(-2,-6),Maximum(0,2),Minimum(1/2 , 29/16)