What is the extreme f(x)=x^3-12x^2-27x+2,-2<= x<= 0 ?

The extreme f(x)=x^3-12x^2-27x+2,-2<= x<= 0 is Minimum(-2,0),Maximum(-1,16),Minimum(0,2)

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extreme f(x)=x^3-12x^2-27x+2,-2<= x<= 0

extreme points

f (x) = x^{3} - 12 x^{2} - 27 x + 2, - 2 \leq x \leq 0

Minimum (- 2, 0), Maximum (- 1, 16), Minimum (0, 2)

Solution steps

Find the critical points:

x = - 2, x = - 1, x = 0

Domain of

x^{3} - 12 x^{2} - 27 x + 2, - 2 \leq x \leq 0 : - 2 \leq x \leq 0

Find intervals:

Increasing

: - 2 < x < - 1,

Decreasing

: - 1 < x < 0

Plug the extreme point

x = - 2

into

x^{3} - 12 x^{2} - 27 x + 2 \Rightarrow y = 0

Minimum (- 2, 0)

Plug the extreme point

x = - 1

into

x^{3} - 12 x^{2} - 27 x + 2 \Rightarrow y = 16

Maximum (- 1, 16)

Plug the extreme point

x = 0

into

x^{3} - 12 x^{2} - 27 x + 2 \Rightarrow y = 2

Minimum (0, 2)

Minimum (- 2, 0), Maximum (- 1, 16), Minimum (0, 2)

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

f (x) = \frac{8 x}{x ^{2} + 16}

\frac{( x ^{2} - 1 )}{( x - 3 ) ^{2}}

s (t) = - 4.9 t^{2} + 29 t + 2

y = (x + 3)^{2} - 7

What is the extreme f(x)=x^3-12x^2-27x+2,-2<= x<= 0 ?
The extreme f(x)=x^3-12x^2-27x+2,-2<= x<= 0 is Minimum(-2,0),Maximum(-1,16),Minimum(0,2)