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

The extreme f(x)=x^2-2x,-1<= x<= 2 is Maximum(-1,3),Minimum(1,-1),Maximum(2,0)

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

extreme points

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

Maximum (- 1, 3), Minimum (1, - 1), Maximum (2, 0)

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = - 1

into

x^{2} - 2 x \Rightarrow y = 3

Maximum (- 1, 3)

Plug the extreme point

x = 1

into

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

Minimum (1, - 1)

Plug the extreme point

x = 2

into

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

Maximum (2, 0)

Maximum (- 1, 3), Minimum (1, - 1), Maximum (2, 0)

f (x) = - x^{3} + 3 x^{2} - 6, - 2 \leq x \leq 5

\frac{2 x - 2}{x ^{2} + x} + 1

f (x, y) = x^{2} + y^{2} - 6 x - 6 y

f (x) = \frac{- 4}{x ^{2} - 9}

f (x, y) = e^{\frac{( x ^{2} + y ^{2} )}{2}}

What is the extreme f(x)=x^2-2x,-1<= x<= 2 ?
The extreme f(x)=x^2-2x,-1<= x<= 2 is Maximum(-1,3),Minimum(1,-1),Maximum(2,0)