What is the extreme f(x)=-x^2+11,-3<= x<= 4 ?

The extreme f(x)=-x^2+11,-3<= x<= 4 is Minimum(-3,2),Maximum(0,11),Minimum(4,-5)

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extreme f(x)=-x^2+11,-3<= x<= 4

extreme points

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

Minimum (- 3, 2), Maximum (0, 11), Minimum (4, - 5)

Solution steps

Find the critical points:

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

Domain of

- x^{2} + 11, - 3 \leq x \leq 4 : - 3 \leq x \leq 4

Find intervals:

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 4

Plug the extreme point

x = - 3

into

- x^{2} + 11 \Rightarrow y = 2

Minimum (- 3, 2)

Plug the extreme point

x = 0

into

- x^{2} + 11 \Rightarrow y = 11

Maximum (0, 11)

Plug the extreme point

x = 4

into

- x^{2} + 11 \Rightarrow y = - 5

Minimum (4, - 5)

Minimum (- 3, 2), Maximum (0, 11), Minimum (4, - 5)

f (x) = sin (x) cos (x) + 2

f (x) = \frac{( x - 2 ) ^{2}}{25} + \frac{( y - 7 ) ^{2}}{36}

f (x) = (\frac{( x ^{2} - 1 ) ^{0.5}}{x ^{3}})

f (x) = x^{2} + x - 3 x y + y^{3} - 5

f (x, y) = 6 x^{2} + 9 x y + 3 y^{2} + 8 x y^{2} + 5 x^{2} y

What is the extreme f(x)=-x^2+11,-3<= x<= 4 ?
The extreme f(x)=-x^2+11,-3<= x<= 4 is Minimum(-3,2),Maximum(0,11),Minimum(4,-5)