What is the extreme y=x^4+2x^3-2x^2+1 ?

The extreme y=x^4+2x^3-2x^2+1 is Minimum(-2,-7),Maximum(0,1),Minimum(1/2 , 13/16)

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extreme y=x^4+2x^3-2x^2+1

extreme points

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

Minimum (- 2, - 7), Maximum (0, 1), Minimum (\frac{1}{2}, \frac{13}{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} + 1 : - 7

Minimum (- 2, - 7)

Plug

x = 0

into

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

Maximum (0, 1)

Plug

x = \frac{1}{2}

into

x^{4} + 2 x^{3} - 2 x^{2} + 1 : \frac{13}{16}

Minimum (\frac{1}{2}, \frac{13}{16})

Minimum (- 2, - 7), Maximum (0, 1), Minimum (\frac{1}{2}, \frac{13}{16})

m = 16, (- 1, - \frac{9}{2})

\frac{x ^{2} + 9 x + 20}{x + 4}

y = x + \frac{3600}{x}

\frac{\frac{4}{x}}{\frac{4}{x} + 4}

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

What is the extreme y=x^4+2x^3-2x^2+1 ?
The extreme y=x^4+2x^3-2x^2+1 is Minimum(-2,-7),Maximum(0,1),Minimum(1/2 , 13/16)