What is the extreme f(x)=-2x^2-1,-4<= x<= 3 ?

The extreme f(x)=-2x^2-1,-4<= x<= 3 is Minimum(-4,-33),Maximum(0,-1),Minimum(3,-19)

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extreme f(x)=-2x^2-1,-4<= x<= 3

extreme points

f (x) = - 2 x^{2} - 1, - 4 \leq x \leq 3

Minimum (- 4, - 33), Maximum (0, - 1), Minimum (3, - 19)

Solution steps

Find the critical points:

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

Domain of

- 2 x^{2} - 1, - 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

- 2 x^{2} - 1 \Rightarrow y = - 33

Minimum (- 4, - 33)

Plug the extreme point

x = 0

into

- 2 x^{2} - 1 \Rightarrow y = - 1

Maximum (0, - 1)

Plug the extreme point

x = 3

into

- 2 x^{2} - 1 \Rightarrow y = - 19

Minimum (3, - 19)

Minimum (- 4, - 33), Maximum (0, - 1), Minimum (3, - 19)

f (x) = \frac{x}{ln ( 2 x )}

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

f (x) = \frac{x ^{2}}{x ^{2} + 3}, - 1 \leq x \leq 1

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

f (x) = \frac{x ^{2}}{x ^{2} + 1}, - 2 \leq x \leq 2

What is the extreme f(x)=-2x^2-1,-4<= x<= 3 ?
The extreme f(x)=-2x^2-1,-4<= x<= 3 is Minimum(-4,-33),Maximum(0,-1),Minimum(3,-19)