What is the extreme f(x)=x^3-9x^2,-1<= x<= 7 ?

The extreme f(x)=x^3-9x^2,-1<= x<= 7 is Minimum(-1,-10),Maximum(0,0),Minimum(6,-108),Maximum(7,-98)

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extreme f(x)=x^3-9x^2,-1<= x<= 7

extreme points

f (x) = x^{3} - 9 x^{2}, - 1 \leq x \leq 7

Minimum (- 1, - 10), Maximum (0, 0), Minimum (6, - 108), Maximum (7, - 98)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 6, x = 7

Domain of

x^{3} - 9 x^{2}, - 1 \leq x \leq 7 : - 1 \leq x \leq 7

Find intervals:

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < 7

Plug the extreme point

x = - 1

into

x^{3} - 9 x^{2} \Rightarrow y = - 10

Minimum (- 1, - 10)

Plug the extreme point

x = 0

into

x^{3} - 9 x^{2} \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 6

into

x^{3} - 9 x^{2} \Rightarrow y = - 108

Minimum (6, - 108)

Plug the extreme point

x = 7

into

x^{3} - 9 x^{2} \Rightarrow y = - 98

Maximum (7, - 98)

Minimum (- 1, - 10), Maximum (0, 0), Minimum (6, - 108), Maximum (7, - 98)

f (x, y) = x^{2} + y^{2} + 9

f (x, y) = x^{2} + y^{2} - 2 x + 2

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

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 6 y + 19

f (x) = x^{3} - 24 x^{2} + 135 x

What is the extreme f(x)=x^3-9x^2,-1<= x<= 7 ?
The extreme f(x)=x^3-9x^2,-1<= x<= 7 is Minimum(-1,-10),Maximum(0,0),Minimum(6,-108),Maximum(7,-98)