What is the extreme f(x)=12+2x-x^2,0<= x<= 5 ?

The extreme f(x)=12+2x-x^2,0<= x<= 5 is Minimum(0,12),Maximum(1,13),Minimum(5,-3)

extreme f(x)=12+2x-x^2,0<= x<= 5

extreme points

f (x) = 12 + 2 x - x^{2}, 0 \leq x \leq 5

Minimum (0, 12), Maximum (1, 13), Minimum (5, - 3)

Solution steps

Find the critical points:

x = 0, x = 1, x = 5

Domain of

12 + 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

12 + 2 x - x^{2} \Rightarrow y = 12

Minimum (0, 12)

Plug the extreme point

x = 1

into

12 + 2 x - x^{2} \Rightarrow y = 13

Maximum (1, 13)

Plug the extreme point

x = 5

into

12 + 2 x - x^{2} \Rightarrow y = - 3

Minimum (5, - 3)

Minimum (0, 12), Maximum (1, 13), Minimum (5, - 3)

f (x) = 1 - x

f (x, y) = x^{2} + x y + y^{2} + 3 x - 3 y + 7

f (x) = x^{2} - 6, (- 2, 4)

B (1 + p)

f (x) = x - 4 ln (x)

What is the extreme f(x)=12+2x-x^2,0<= x<= 5 ?
The extreme f(x)=12+2x-x^2,0<= x<= 5 is Minimum(0,12),Maximum(1,13),Minimum(5,-3)