What is the extreme f(x)=3x^3+8,-2<= x<= 2 ?

The extreme f(x)=3x^3+8,-2<= x<= 2 is Minimum(-2,-16),Saddle(0,8),Maximum(2,32)

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extreme f(x)=3x^3+8,-2<= x<= 2

extreme points

f (x) = 3 x^{3} + 8, - 2 \leq x \leq 2

Minimum (- 2, - 16), Saddle (0, 8), Maximum (2, 32)

Solution steps

Find the critical points:

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

Domain of

3 x^{3} + 8, - 2 \leq x \leq 2 : - 2 \leq x \leq 2

Find intervals:

Increasing

: - 2 < x < 0,

Increasing

: 0 < x < 2

Plug the extreme point

x = - 2

into

3 x^{3} + 8 \Rightarrow y = - 16

Minimum (- 2, - 16)

Plug the extreme point

x = 0

into

3 x^{3} + 8 \Rightarrow y = 8

Saddle (0, 8)

Plug the extreme point

x = 2

into

3 x^{3} + 8 \Rightarrow y = 32

Maximum (2, 32)

Minimum (- 2, - 16), Saddle (0, 8), Maximum (2, 32)

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

f (x, y) = y + x

\frac{( 4 x ^{2} )}{x ^{2} + 3}

f (x) = 3 x^{2} - 24 x

x y + \frac{1}{x} + \frac{1}{y}

What is the extreme f(x)=3x^3+8,-2<= x<= 2 ?
The extreme f(x)=3x^3+8,-2<= x<= 2 is Minimum(-2,-16),Saddle(0,8),Maximum(2,32)