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

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

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

extreme points

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

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

Solution steps

Find the critical points:

x = - 2, x = 0, x = 3

Domain of

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

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 3

Plug the extreme point

x = - 2

into

- x^{2} + 4 \Rightarrow y = 0

Minimum (- 2, 0)

Plug the extreme point

x = 0

into

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

Maximum (0, 4)

Plug the extreme point

x = 3

into

- x^{2} + 4 \Rightarrow y = - 5

Minimum (3, - 5)

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

x^{\frac{1}{5}} (x + 6)

f (x) = x^{2} e^{4 x}

f (x) = (x + 1)^{5} - 5 x - 2

f (x) = 5 x^{5} - 3 x^{3}

f (x) = \frac{- 5 x ^{2} + 16 x}{2 4 - x}

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