What is the extreme f(x)=-3x^2-5,-4<= x<= 3 ?

The extreme f(x)=-3x^2-5,-4<= x<= 3 is Minimum(-4,-53),Maximum(0,-5),Minimum(3,-32)

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extreme f(x)=-3x^2-5,-4<= x<= 3

extreme points

f (x) = - 3 x^{2} - 5, - 4 \leq x \leq 3

Minimum (- 4, - 53), Maximum (0, - 5), Minimum (3, - 32)

Solution steps

Find the critical points:

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

Domain of

- 3 x^{2} - 5, - 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

- 3 x^{2} - 5 \Rightarrow y = - 53

Minimum (- 4, - 53)

Plug the extreme point

x = 0

into

- 3 x^{2} - 5 \Rightarrow y = - 5

Maximum (0, - 5)

Plug the extreme point

x = 3

into

- 3 x^{2} - 5 \Rightarrow y = - 32

Minimum (3, - 32)

Minimum (- 4, - 53), Maximum (0, - 5), Minimum (3, - 32)

f (x, y) = x^{2} + y^{2} + 8 x - 10 y

f (x) = ln (2 x^{2} - 2 x + 3), 0 \leq x \leq 2

f (x) = 5 - 6 x^{2}, - 4 \leq x \leq 2

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

e^{9 x} + e^{- x}

What is the extreme f(x)=-3x^2-5,-4<= x<= 3 ?
The extreme f(x)=-3x^2-5,-4<= x<= 3 is Minimum(-4,-53),Maximum(0,-5),Minimum(3,-32)