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

The extreme f(x)=x^2-x-3,0<= x<= 3 is Maximum(0,-3),Minimum(1/2 ,-13/4),Maximum(3,3)

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

extreme points

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

Maximum (0, - 3), Minimum (\frac{1}{2}, - \frac{13}{4}), Maximum (3, 3)

Solution steps

Find the critical points:

x = 0, x = \frac{1}{2}, x = 3

Domain of

x^{2} - x - 3, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Decreasing

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

Increasing

: \frac{1}{2} < x < 3

Plug the extreme point

x = 0

into

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

Maximum (0, - 3)

Plug the extreme point

x = \frac{1}{2}

into

x^{2} - x - 3 \Rightarrow y = - \frac{13}{4}

Minimum (\frac{1}{2}, - \frac{13}{4})

Plug the extreme point

x = 3

into

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

Maximum (3, 3)

Maximum (0, - 3), Minimum (\frac{1}{2}, - \frac{13}{4}), Maximum (3, 3)

f (x) = - 4 x^{6} ln (3 x)

f (x) = - x^{2} + 100 x + 8

f (x) = - x^{3} + 4 x^{2} + 16 x - 7

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

f (x) = - 7 x^{3} + 21 x + 4

What is the extreme f(x)=x^2-x-3,0<= x<= 3 ?
The extreme f(x)=x^2-x-3,0<= x<= 3 is Maximum(0,-3),Minimum(1/2 ,-13/4),Maximum(3,3)