What is the extreme f(x)=5+6x-x^2,0<= x<= 4 ?

The extreme f(x)=5+6x-x^2,0<= x<= 4 is Minimum(0,5),Maximum(3,14),Minimum(4,13)

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extreme f(x)=5+6x-x^2,0<= x<= 4

extreme points

f (x) = 5 + 6 x - x^{2}, 0 \leq x \leq 4

Minimum (0, 5), Maximum (3, 14), Minimum (4, 13)

Solution steps

Find the critical points:

x = 0, x = 3, x = 4

Domain of

5 + 6 x - x^{2}, 0 \leq x \leq 4 : 0 \leq x \leq 4

Find intervals:

Increasing

: 0 < x < 3,

Decreasing

: 3 < x < 4

Plug the extreme point

x = 0

into

5 + 6 x - x^{2} \Rightarrow y = 5

Minimum (0, 5)

Plug the extreme point

x = 3

into

5 + 6 x - x^{2} \Rightarrow y = 14

Maximum (3, 14)

Plug the extreme point

x = 4

into

5 + 6 x - x^{2} \Rightarrow y = 13

Minimum (4, 13)

Minimum (0, 5), Maximum (3, 14), Minimum (4, 13)

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

f (x) = - x^{3} + 8 x^{2} - 15 x

y = 2 x^{3} - 15 x^{2} + 24 x - 5

f (x) = - 7 x^{2} + 42 x + 5

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

What is the extreme f(x)=5+6x-x^2,0<= x<= 4 ?
The extreme f(x)=5+6x-x^2,0<= x<= 4 is Minimum(0,5),Maximum(3,14),Minimum(4,13)