What is the extreme f(x)=-x^2+6x+1,5<= x<= 8 ?

The extreme f(x)=-x^2+6x+1,5<= x<= 8 is Maximum(5,6),Minimum(8,-15)

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extreme f(x)=-x^2+6x+1,5<= x<= 8

extreme points

f (x) = - x^{2} + 6 x + 1, 5 \leq x \leq 8

Maximum (5, 6), Minimum (8, - 15)

Solution steps

Find the critical points:

x = 5, x = 8

Domain of

- x^{2} + 6 x + 1, 5 \leq x \leq 8 : 5 \leq x \leq 8

Find intervals:

Decreasing

: 5 < x < 8

Plug the extreme point

x = 5

into

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

Maximum (5, 6)

Plug the extreme point

x = 8

into

- x^{2} + 6 x + 1 \Rightarrow y = - 15

Minimum (8, - 15)

Maximum (5, 6), Minimum (8, - 15)

f (x) = x^{3} + 2 x^{2} - 4 x + 8

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

f (x, y) = x^{2} + y^{2} - 81

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

f (x) = 3 sin (2 x)

What is the extreme f(x)=-x^2+6x+1,5<= x<= 8 ?
The extreme f(x)=-x^2+6x+1,5<= x<= 8 is Maximum(5,6),Minimum(8,-15)