What is the extreme f(x)=12+4x-x^2,0<= x<= 5 ?

The extreme f(x)=12+4x-x^2,0<= x<= 5 is Minimum(0,12),Maximum(2,16),Minimum(5,7)

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extreme f(x)=12+4x-x^2,0<= x<= 5

extreme points

f (x) = 12 + 4 x - x^{2}, 0 \leq x \leq 5

Minimum (0, 12), Maximum (2, 16), Minimum (5, 7)

Solution steps

Find the critical points:

x = 0, x = 2, x = 5

Domain of

12 + 4 x - x^{2}, 0 \leq x \leq 5 : 0 \leq x \leq 5

Find intervals:

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 5

Plug the extreme point

x = 0

into

12 + 4 x - x^{2} \Rightarrow y = 12

Minimum (0, 12)

Plug the extreme point

x = 2

into

12 + 4 x - x^{2} \Rightarrow y = 16

Maximum (2, 16)

Plug the extreme point

x = 5

into

12 + 4 x - x^{2} \Rightarrow y = 7

Minimum (5, 7)

Minimum (0, 12), Maximum (2, 16), Minimum (5, 7)

f (x) = 6

6 x - 10 y = - 3

f (x) = 7 - 2 x

f (x) = (x - 3)

\frac{1}{x ^{2} + 5 x - 24}

What is the extreme f(x)=12+4x-x^2,0<= x<= 5 ?
The extreme f(x)=12+4x-x^2,0<= x<= 5 is Minimum(0,12),Maximum(2,16),Minimum(5,7)