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

The extreme f(x)=x^2-x-2,0<= x<= 2 is Maximum(0,-2),Minimum(1/2 ,-9/4),Maximum(2,0)

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

extreme points

f (x) = x^{2} - x - 2, 0 \leq x \leq 2

Maximum (0, - 2), Minimum (\frac{1}{2}, - \frac{9}{4}), Maximum (2, 0)

Solution steps

Find the critical points:

x = 0, x = \frac{1}{2}, x = 2

Domain of

x^{2} - x - 2, 0 \leq x \leq 2 : 0 \leq x \leq 2

Find intervals:

Decreasing

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

Increasing

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

Plug the extreme point

x = 0

into

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

Maximum (0, - 2)

Plug the extreme point

x = \frac{1}{2}

into

x^{2} - x - 2 \Rightarrow y = - \frac{9}{4}

Minimum (\frac{1}{2}, - \frac{9}{4})

Plug the extreme point

x = 2

into

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

Maximum (2, 0)

Maximum (0, - 2), Minimum (\frac{1}{2}, - \frac{9}{4}), Maximum (2, 0)

- x^{3} + 12 x - 19

f (x) = 68 - x^{2} - y^{2}

y = \frac{x ^{2}}{x ^{2} + 243}

f (x, y) = ln (2 x^{2} + y^{2} - 1)

f (x) = 2 x^{2} - 8 x y + 10 y^{2} + 10 x - 4 y - 5

What is the extreme f(x)=x^2-x-2,0<= x<= 2 ?
The extreme f(x)=x^2-x-2,0<= x<= 2 is Maximum(0,-2),Minimum(1/2 ,-9/4),Maximum(2,0)