What is the extreme f(x)=17+2x-x^2,0<= x<= 5 ?

The extreme f(x)=17+2x-x^2,0<= x<= 5 is Minimum(0,17),Maximum(1,18),Minimum(5,2)

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extreme f(x)=17+2x-x^2,0<= x<= 5

extreme points

f (x) = 17 + 2 x - x^{2}, 0 \leq x \leq 5

Minimum (0, 17), Maximum (1, 18), Minimum (5, 2)

Solution steps

Find the critical points:

x = 0, x = 1, x = 5

Domain of

17 + 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

17 + 2 x - x^{2} \Rightarrow y = 17

Minimum (0, 17)

Plug the extreme point

x = 1

into

17 + 2 x - x^{2} \Rightarrow y = 18

Maximum (1, 18)

Plug the extreme point

x = 5

into

17 + 2 x - x^{2} \Rightarrow y = 2

Minimum (5, 2)

Minimum (0, 17), Maximum (1, 18), Minimum (5, 2)

f (x) = 3 x^{2} - 36 x + 24

f (x) = 9 x^{2} - 3 y^{2}

f (x) = 25 x - 50 sin (x), (- \frac{π}{4}, \frac{π}{2})

f (x, y) = - 6 x^{2} + 5 y^{2} + 12 x - 20 y - 18

x - x

What is the extreme f(x)=17+2x-x^2,0<= x<= 5 ?
The extreme f(x)=17+2x-x^2,0<= x<= 5 is Minimum(0,17),Maximum(1,18),Minimum(5,2)