What is the extreme f(x)=x^3+6x^2-63x+5,-8<= x<= 0 ?

The extreme f(x)=x^3+6x^2-63x+5,-8<= x<= 0 is Minimum(-8,381),Maximum(-7,397),Minimum(0,5)

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extreme f(x)=x^3+6x^2-63x+5,-8<= x<= 0

extreme points

f (x) = x^{3} + 6 x^{2} - 63 x + 5, - 8 \leq x \leq 0

Minimum (- 8, 381), Maximum (- 7, 397), Minimum (0, 5)

Solution steps

Find the critical points:

x = - 8, x = - 7, x = 0

Domain of

x^{3} + 6 x^{2} - 63 x + 5, - 8 \leq x \leq 0 : - 8 \leq x \leq 0

Find intervals:

Increasing

: - 8 < x < - 7,

Decreasing

: - 7 < x < 0

Plug the extreme point

x = - 8

into

x^{3} + 6 x^{2} - 63 x + 5 \Rightarrow y = 381

Minimum (- 8, 381)

Plug the extreme point

x = - 7

into

x^{3} + 6 x^{2} - 63 x + 5 \Rightarrow y = 397

Maximum (- 7, 397)

Plug the extreme point

x = 0

into

x^{3} + 6 x^{2} - 63 x + 5 \Rightarrow y = 5

Minimum (0, 5)

Minimum (- 8, 381), Maximum (- 7, 397), Minimum (0, 5)

f (x) = 3 x^{2} - 6 x + 24

f (x) = (9 x^{2} + 4 x + 5) (9 y^{2} + 6 y + 6)

f (x) = \frac{4860}{x} + 12 x + 709407

f (x) = x^{3} + 6 x^{2} - 63 x + 12, - 5 \leq x \leq 4

9 sin (x) + 9 cos (x), 0 \leq x \leq 2 π

What is the extreme f(x)=x^3+6x^2-63x+5,-8<= x<= 0 ?
The extreme f(x)=x^3+6x^2-63x+5,-8<= x<= 0 is Minimum(-8,381),Maximum(-7,397),Minimum(0,5)