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

The extreme f(x)=2-2x^2,-4<= x<= 1 is Minimum(-4,-30),Maximum(0,2),Minimum(1,0)

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

extreme points

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

Minimum (- 4, - 30), Maximum (0, 2), Minimum (1, 0)

Solution steps

Find the critical points:

x = - 4, x = 0, x = 1

Domain of

2 - 2 x^{2}, - 4 \leq x \leq 1 : - 4 \leq x \leq 1

Find intervals:

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 1

Plug the extreme point

x = - 4

into

2 - 2 x^{2} \Rightarrow y = - 30

Minimum (- 4, - 30)

Plug the extreme point

x = 0

into

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

Maximum (0, 2)

Plug the extreme point

x = 1

into

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

Minimum (1, 0)

Minimum (- 4, - 30), Maximum (0, 2), Minimum (1, 0)

f (x) = 3 x + sin (3 x)

f (x) = 6 x^{3} - 9 x^{2} - 216 x + 7, - 4 \leq x \leq 5

e^{x^{2} + 3 \cdot x + 4}

f (x) = x^{4} + 3 x^{3} - 9 x^{2} - 23 x - 12

f (x) = x^{3} - 4 x^{2} - 3 x^{2} + 6

What is the extreme f(x)=2-2x^2,-4<= x<= 1 ?
The extreme f(x)=2-2x^2,-4<= x<= 1 is Minimum(-4,-30),Maximum(0,2),Minimum(1,0)