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

The extreme f(x)=16+4x-x^2,0<= x<= 5 is Minimum(0,16),Maximum(2,20),Minimum(5,11)

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extreme f(x)=16+4x-x^2,0<= x<= 5

extreme points

f (x) = 16 + 4 x - x^{2}, 0 \leq x \leq 5

Minimum (0, 16), Maximum (2, 20), Minimum (5, 11)

Solution steps

Find the critical points:

x = 0, x = 2, x = 5

Domain of

16 + 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

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

Minimum (0, 16)

Plug the extreme point

x = 2

into

16 + 4 x - x^{2} \Rightarrow y = 20

Maximum (2, 20)

Plug the extreme point

x = 5

into

16 + 4 x - x^{2} \Rightarrow y = 11

Minimum (5, 11)

Minimum (0, 16), Maximum (2, 20), Minimum (5, 11)

f (x) = \frac{x ^{3}}{3} - x^{2} - 3 x - 1

f (x, y) = x^{4} + y^{4} - 16 x y

f (x) = x e^{\frac{- x ^{2}}{162}}

f (x) = \frac{x ^{4}}{4} - \frac{3 x ^{2}}{2}

What is the extreme f(x)=16+4x-x^2,0<= x<= 5 ?
The extreme f(x)=16+4x-x^2,0<= x<= 5 is Minimum(0,16),Maximum(2,20),Minimum(5,11)