What is the extreme f(x)=2x-3,0<= x<= 2 ?

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

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extreme f(x)=2x-3,0<= x<= 2

extreme points

f (x) = 2 x - 3, 0 \leq x \leq 2

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

Solution steps

Find the critical points:

x = 0, x = 2

Domain of

2 x - 3, 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Increasing

: 0 < x < 2

Plug the extreme point

x = 0

into

2 x - 3 \Rightarrow y = - 3

Minimum (0, - 3)

Plug the extreme point

x = 2

into

2 x - 3 \Rightarrow y = 1

Maximum (2, 1)

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

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

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

x (r, t) = 2 r (2 t)

\frac{1}{4 x ^{2}} - \frac{1}{x ^{3}}

- x^{2} + 4 x y - 5 y^{2} - 6 x + 22 y + 9

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