What is the extreme f(x)=x^2-6x-1,0<= x<= 3 ?

The extreme f(x)=x^2-6x-1,0<= x<= 3 is Maximum(0,-1),Minimum(3,-10)

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extreme f(x)=x^2-6x-1,0<= x<= 3

extreme points

f (x) = x^{2} - 6 x - 1, 0 \leq x \leq 3

Maximum (0, - 1), Minimum (3, - 10)

Solution steps

Find the critical points:

x = 0, x = 3

Domain of

x^{2} - 6 x - 1, 0 \leq x \leq 3 : 0 \leq x \leq 3

Find intervals:

Decreasing

: 0 < x < 3

Plug the extreme point

x = 0

into

x^{2} - 6 x - 1 \Rightarrow y = - 1

Maximum (0, - 1)

Plug the extreme point

x = 3

into

x^{2} - 6 x - 1 \Rightarrow y = - 10

Minimum (3, - 10)

Maximum (0, - 1), Minimum (3, - 10)

f (x) = x^{2} - 6 x - 1, 0 \leq x \leq 5

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

y = \frac{2 x + 5 z + 19}{3}

g (x) = x^{3} - x^{2} - 8 x - 5

f (x) = \frac{x ^{2} + 1}{x ^{2} - 25}

What is the extreme f(x)=x^2-6x-1,0<= x<= 3 ?
The extreme f(x)=x^2-6x-1,0<= x<= 3 is Maximum(0,-1),Minimum(3,-10)