What is the extreme f(x)=13+2x-x^2,0<= x<= 5 ?

The extreme f(x)=13+2x-x^2,0<= x<= 5 is Minimum(0,13),Maximum(1,14),Minimum(5,-2)

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extreme f(x)=13+2x-x^2,0<= x<= 5

extreme points

f (x) = 13 + 2 x - x^{2}, 0 \leq x \leq 5

Minimum (0, 13), Maximum (1, 14), Minimum (5, - 2)

Solution steps

Find the critical points:

x = 0, x = 1, x = 5

Domain of

13 + 2 x - x^{2}, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 5

Plug the extreme point

x = 0

into

13 + 2 x - x^{2} \Rightarrow y = 13

Minimum (0, 13)

Plug the extreme point

x = 1

into

13 + 2 x - x^{2} \Rightarrow y = 14

Maximum (1, 14)

Plug the extreme point

x = 5

into

13 + 2 x - x^{2} \Rightarrow y = - 2

Minimum (5, - 2)

Minimum (0, 13), Maximum (1, 14), Minimum (5, - 2)

f (x) = x^{4} - 18 x^{2} + 5

f (x) = (x^{3} - 4 x y) e^{- 2 y}

f (x) = - 2 (x - 3) (x + 5)

x^{3} - 2 x^{2} - 15 x + 5, - 2 \leq x \leq 0

x^{2} + y x + y^{2}

What is the extreme f(x)=13+2x-x^2,0<= x<= 5 ?
The extreme f(x)=13+2x-x^2,0<= x<= 5 is Minimum(0,13),Maximum(1,14),Minimum(5,-2)