What is the extreme f(x)=19+4x-x^2,0<= x<= 5 ?

The extreme f(x)=19+4x-x^2,0<= x<= 5 is Minimum(0,19),Maximum(2,23),Minimum(5,14)

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extreme f(x)=19+4x-x^2,0<= x<= 5

extreme points

f (x) = 19 + 4 x - x^{2}, 0 \leq x \leq 5

Minimum (0, 19), Maximum (2, 23), Minimum (5, 14)

Solution steps

Find the critical points:

x = 0, x = 2, x = 5

Domain of

19 + 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

19 + 4 x - x^{2} \Rightarrow y = 19

Minimum (0, 19)

Plug the extreme point

x = 2

into

19 + 4 x - x^{2} \Rightarrow y = 23

Maximum (2, 23)

Plug the extreme point

x = 5

into

19 + 4 x - x^{2} \Rightarrow y = 14

Minimum (5, 14)

Minimum (0, 19), Maximum (2, 23), Minimum (5, 14)

x^{2} + 3 x - 7

x^{2} + 3 x - 4

f (x) = 6 x^{4} - 8 x^{3} + 2

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

f (x) = 2 x - \frac{32}{x}

What is the extreme f(x)=19+4x-x^2,0<= x<= 5 ?
The extreme f(x)=19+4x-x^2,0<= x<= 5 is Minimum(0,19),Maximum(2,23),Minimum(5,14)