What is the extreme f(x)=8-x,-1<= x<= 3 ?

The extreme f(x)=8-x,-1<= x<= 3 is Maximum(-1,9),Minimum(3,5)

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extreme f(x)=8-x,-1<= x<= 3

extreme points

f (x) = 8 - x, - 1 \leq x \leq 3

Maximum (- 1, 9), Minimum (3, 5)

Solution steps

Find the critical points:

x = - 1, x = 3

Domain of

8 - x, - 1 \leq x \leq 3 : - 1 \leq x \leq 3

Find intervals:

Decreasing

: - 1 < x < 3

Plug the extreme point

x = - 1

into

8 - x \Rightarrow y = 9

Maximum (- 1, 9)

Plug the extreme point

x = 3

into

8 - x \Rightarrow y = 5

Minimum (3, 5)

Maximum (- 1, 9), Minimum (3, 5)

P (x, y) = x^{2} + 5 x y - 4 y^{2}

f (x) = 2 x^{3} - 6 x^{2} - 18 x + 2

f (x) = cos (3 x), - \frac{1}{6} π < x < \frac{1}{2} π

f (x) = 8 x^{3}

f (x y) = 2 x y + 80 y - 15 y^{2} - 0.1 x^{2} y - 120

What is the extreme f(x)=8-x,-1<= x<= 3 ?
The extreme f(x)=8-x,-1<= x<= 3 is Maximum(-1,9),Minimum(3,5)