What is the extreme f(x)=-x^2+8,-2<= x<= 4 ?

The extreme f(x)=-x^2+8,-2<= x<= 4 is Minimum(-2,4),Maximum(0,8),Minimum(4,-8)

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extreme f(x)=-x^2+8,-2<= x<= 4

extreme points

f (x) = - x^{2} + 8, - 2 \leq x \leq 4

Minimum (- 2, 4), Maximum (0, 8), Minimum (4, - 8)

Solution steps

Find the critical points:

x = - 2, x = 0, x = 4

Domain of

- x^{2} + 8, - 2 \leq x \leq 4 : - 2 \leq x \leq 4

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 4

Plug the extreme point

x = - 2

into

- x^{2} + 8 \Rightarrow y = 4

Minimum (- 2, 4)

Plug the extreme point

x = 0

into

- x^{2} + 8 \Rightarrow y = 8

Maximum (0, 8)

Plug the extreme point

x = 4

into

- x^{2} + 8 \Rightarrow y = - 8

Minimum (4, - 8)

Minimum (- 2, 4), Maximum (0, 8), Minimum (4, - 8)

f (x) = - x^{4} + 4 x^{3} + 9 x - 6

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

f (x, y) = - 3 x^{2} - 6 y^{2} - 6 x y + 66 x + 102 y

89 e^{x^{4}}

- 4.9 x^{2} + 10 x + 3

What is the extreme f(x)=-x^2+8,-2<= x<= 4 ?
The extreme f(x)=-x^2+8,-2<= x<= 4 is Minimum(-2,4),Maximum(0,8),Minimum(4,-8)