What is the extreme f(x)=x^3-6x^2+8,-1<= x<= 6 ?

The extreme f(x)=x^3-6x^2+8,-1<= x<= 6 is Minimum(-1,1),Maximum(0,8),Minimum(4,-24),Maximum(6,8)

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extreme f(x)=x^3-6x^2+8,-1<= x<= 6

extreme points

f (x) = x^{3} - 6 x^{2} + 8, - 1 \leq x \leq 6

Minimum (- 1, 1), Maximum (0, 8), Minimum (4, - 24), Maximum (6, 8)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 4, x = 6

Domain of

x^{3} - 6 x^{2} + 8, - 1 \leq x \leq 6 : - 1 \leq x \leq 6

Find intervals:

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < 6

Plug the extreme point

x = - 1

into

x^{3} - 6 x^{2} + 8 \Rightarrow y = 1

Minimum (- 1, 1)

Plug the extreme point

x = 0

into

x^{3} - 6 x^{2} + 8 \Rightarrow y = 8

Maximum (0, 8)

Plug the extreme point

x = 4

into

x^{3} - 6 x^{2} + 8 \Rightarrow y = - 24

Minimum (4, - 24)

Plug the extreme point

x = 6

into

x^{3} - 6 x^{2} + 8 \Rightarrow y = 8

Maximum (6, 8)

Minimum (- 1, 1), Maximum (0, 8), Minimum (4, - 24), Maximum (6, 8)

f (x, y) = x^{2} - x y + y^{2} + 5

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

\frac{x ^{2} - 2 x - 3}{x ^{2}}

f (x) = \frac{- 4 t ^{2} + 50 t + 16}{( t ^{2} + 4 ) ^{2}}

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

What is the extreme f(x)=x^3-6x^2+8,-1<= x<= 6 ?
The extreme f(x)=x^3-6x^2+8,-1<= x<= 6 is Minimum(-1,1),Maximum(0,8),Minimum(4,-24),Maximum(6,8)