What is the extreme f(x)=x^2-2x,1<= x<= 4 ?

The extreme f(x)=x^2-2x,1<= x<= 4 is Minimum(1,-1),Maximum(4,8)

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extreme f(x)=x^2-2x,1<= x<= 4

extreme points

f (x) = x^{2} - 2 x, 1 \leq x \leq 4

Minimum (1, - 1), Maximum (4, 8)

Solution steps

Find the critical points:

x = 1, x = 4

Domain of

x^{2} - 2 x, 1 \leq x \leq 4 : 1 \leq x \leq 4

Find intervals:

Increasing

: 1 < x < 4

Plug the extreme point

x = 1

into

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

Minimum (1, - 1)

Plug the extreme point

x = 4

into

x^{2} - 2 x \Rightarrow y = 8

Maximum (4, 8)

Minimum (1, - 1), Maximum (4, 8)

f (x) = \frac{7 x}{x - 4}

21 x^{5} + 35 x^{4}

f (x) = \frac{x + 1}{x ^{2} - 8 x - 9}

f (x, y) = x^{2} + 4 y^{2} + 6

f (x) = - (x^{2} + 4 x - 4) (x + 2) (x - 1)

What is the extreme f(x)=x^2-2x,1<= x<= 4 ?
The extreme f(x)=x^2-2x,1<= x<= 4 is Minimum(1,-1),Maximum(4,8)