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

The extreme f(x)=-2x^2+2x,-3<= x<= 2 is Minimum(-3,-24),Maximum(1/2 , 1/2),Minimum(2,-4)

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extreme f(x)=-2x^2+2x,-3<= x<= 2

extreme

Minimum (- 3, - 24), Maximum (\frac{1}{2}, \frac{1}{2}), Minimum (2, - 4)

Solution steps

Find the critical points:

x = - 3, x = \frac{1}{2}, x = 2

Domain of

- 2 x^{2} + 2 x, - 3 \leq x \leq 2 : - 3 \leq x \leq 2

Find intervals:

Increasing

: - 3 < x < \frac{1}{2},

Decreasing

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

Plug the extreme point

x = - 3

into

- 2 x^{2} + 2 x \Rightarrow y = - 24

Minimum (- 3, - 24)

Plug the extreme point

x = \frac{1}{2}

into

- 2 x^{2} + 2 x \Rightarrow y = \frac{1}{2}

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

Plug the extreme point

x = 2

into

- 2 x^{2} + 2 x \Rightarrow y = - 4

Minimum (2, - 4)

Minimum (- 3, - 24), Maximum (\frac{1}{2}, \frac{1}{2}), Minimum (2, - 4)

What is the extreme f(x)=-2x^2+2x,-3<= x<= 2 ?
The extreme f(x)=-2x^2+2x,-3<= x<= 2 is Minimum(-3,-24),Maximum(1/2 , 1/2),Minimum(2,-4)