What is the extreme f(x)=-x^2+9,-4<= x<= 3 ?

The extreme f(x)=-x^2+9,-4<= x<= 3 is Minimum(-4,-7),Maximum(0,9),Minimum(3,0)

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extreme f(x)=-x^2+9,-4<= x<= 3

extreme points

f (x) = - x^{2} + 9, - 4 \leq x \leq 3

Minimum (- 4, - 7), Maximum (0, 9), Minimum (3, 0)

Solution steps

Find the critical points:

x = - 4, x = 0, x = 3

Domain of

- x^{2} + 9, - 4 \leq x \leq 3 : - 4 \leq x \leq 3

Find intervals:

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 3

Plug the extreme point

x = - 4

into

- x^{2} + 9 \Rightarrow y = - 7

Minimum (- 4, - 7)

Plug the extreme point

x = 0

into

- x^{2} + 9 \Rightarrow y = 9

Maximum (0, 9)

Plug the extreme point

x = 3

into

- x^{2} + 9 \Rightarrow y = 0

Minimum (3, 0)

Minimum (- 4, - 7), Maximum (0, 9), Minimum (3, 0)

f (x, y) = \frac{x ^{3} + 9}{x ^{2}}

f (x) = x^{2} + 10 x y + y^{2}

f (x) = 14 + 4 x - x^{2}

f (x, y) = x^{3} + x y^{2} + y^{2} - 4 y + 11

8 + 2 x - 2 x^{2}

What is the extreme f(x)=-x^2+9,-4<= x<= 3 ?
The extreme f(x)=-x^2+9,-4<= x<= 3 is Minimum(-4,-7),Maximum(0,9),Minimum(3,0)