What is the extreme f(x)=-x^2+8x-15,3<= x<= 5 ?

The extreme f(x)=-x^2+8x-15,3<= x<= 5 is Minimum(3,0),Maximum(4,1),Minimum(5,0)

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extreme f(x)=-x^2+8x-15,3<= x<= 5

extreme

Minimum (3, 0), Maximum (4, 1), Minimum (5, 0)

Solution steps

Find the critical points:

x = 3, x = 4, x = 5

Domain of

- x^{2} + 8 x - 15, 3 \leq x \leq 5 : 3 \leq x \leq 5

Find intervals:

Increasing

: 3 < x < 4,

Decreasing

: 4 < x < 5

Plug the extreme point

x = 3

into

- x^{2} + 8 x - 15 \Rightarrow y = 0

Minimum (3, 0)

Plug the extreme point

x = 4

into

- x^{2} + 8 x - 15 \Rightarrow y = 1

Maximum (4, 1)

Plug the extreme point

x = 5

into

- x^{2} + 8 x - 15 \Rightarrow y = 0

Minimum (5, 0)

Minimum (3, 0), Maximum (4, 1), Minimum (5, 0)

What is the extreme f(x)=-x^2+8x-15,3<= x<= 5 ?
The extreme f(x)=-x^2+8x-15,3<= x<= 5 is Minimum(3,0),Maximum(4,1),Minimum(5,0)