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

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

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

extreme

Minimum (- 2, - 2), Maximum (0, 0), Minimum (1, - 2)

Solution steps

Find the critical points:

x = - 2, x = 0, x = 1

Domain of

\frac{2 x ^{2}}{x - 2}, - 2 \leq x \leq 1 : - 2 \leq x \leq 1

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 1

Plug the extreme point

x = - 2

into

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

Minimum (- 2, - 2)

Plug the extreme point

x = 0

into

\frac{2 x ^{2}}{x - 2} \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 1

into

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

Minimum (1, - 2)

Minimum (- 2, - 2), Maximum (0, 0), Minimum (1, - 2)

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