What is the extreme (x-x^2)^2,-1<= x<= 1 ?

The extreme (x-x^2)^2,-1<= x<= 1 is Maximum(-1,4),Minimum(0,0),Maximum(1/2 , 1/16),Minimum(1,0)

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extreme (x-x^2)^2,-1<= x<= 1

extreme points

(x - x^{2})^{2}, - 1 \leq x \leq 1

Maximum (- 1, 4), Minimum (0, 0), Maximum (\frac{1}{2}, \frac{1}{16}), Minimum (1, 0)

Solution steps

Find the critical points:

x = - 1, x = 0, x = \frac{1}{2}, x = 1

Domain of

(x - x^{2})^{2}, - 1 \leq x \leq 1 : - 1 \leq x \leq 1

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

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

Decreasing

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

Plug the extreme point

x = - 1

into

(x - x^{2})^{2} \Rightarrow y = 4

Maximum (- 1, 4)

Plug the extreme point

x = 0

into

(x - x^{2})^{2} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = \frac{1}{2}

into

(x - x^{2})^{2} \Rightarrow y = \frac{1}{16}

Maximum (\frac{1}{2}, \frac{1}{16})

Plug the extreme point

x = 1

into

(x - x^{2})^{2} \Rightarrow y = 0

Minimum (1, 0)

Maximum (- 1, 4), Minimum (0, 0), Maximum (\frac{1}{2}, \frac{1}{16}), Minimum (1, 0)

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

f (x, y) = 4 x y

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

f (y, z) = 6 yz - 2 y^{2} z - 3 y z^{2}

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

What is the extreme (x-x^2)^2,-1<= x<= 1 ?
The extreme (x-x^2)^2,-1<= x<= 1 is Maximum(-1,4),Minimum(0,0),Maximum(1/2 , 1/16),Minimum(1,0)