What is the extreme f(x)=x^3+12x^2-27x+6,-10<= x<= 0 ?

The extreme f(x)=x^3+12x^2-27x+6,-10<= x<= 0 is Minimum(-10,476),Maximum(-9,492),Minimum(0,6)

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extreme f(x)=x^3+12x^2-27x+6,-10<= x<= 0

extreme points

f (x) = x^{3} + 12 x^{2} - 27 x + 6, - 10 \leq x \leq 0

Minimum (- 10, 476), Maximum (- 9, 492), Minimum (0, 6)

Solution steps

Find the critical points:

x = - 10, x = - 9, x = 0

Domain of

x^{3} + 12 x^{2} - 27 x + 6, - 10 \leq x \leq 0 : - 10 \leq x \leq 0

Find intervals:

Increasing

: - 10 < x < - 9,

Decreasing

: - 9 < x < 0

Plug the extreme point

x = - 10

into

x^{3} + 12 x^{2} - 27 x + 6 \Rightarrow y = 476

Minimum (- 10, 476)

Plug the extreme point

x = - 9

into

x^{3} + 12 x^{2} - 27 x + 6 \Rightarrow y = 492

Maximum (- 9, 492)

Plug the extreme point

x = 0

into

x^{3} + 12 x^{2} - 27 x + 6 \Rightarrow y = 6

Minimum (0, 6)

Minimum (- 10, 476), Maximum (- 9, 492), Minimum (0, 6)

y = (2 x^{2} - 1)^{2}

f (x) = \frac{6}{- 4 x + 2}

f (y) = x^{y + 1}

f (x) = x^{2} lo g_{8} (x)

x + \frac{192}{x}

What is the extreme f(x)=x^3+12x^2-27x+6,-10<= x<= 0 ?
The extreme f(x)=x^3+12x^2-27x+6,-10<= x<= 0 is Minimum(-10,476),Maximum(-9,492),Minimum(0,6)