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

The extreme f(x)=-x^3+3x^2-6,-2<= x<= 5 is Maximum(-2,14),Minimum(0,-6),Maximum(2,-2),Minimum(5,-56)

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

extreme points

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

Maximum (- 2, 14), Minimum (0, - 6), Maximum (2, - 2), Minimum (5, - 56)

Solution steps

Find the critical points:

x = - 2, x = 0, x = 2, x = 5

Domain of

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

Find intervals:

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 5

Plug the extreme point

x = - 2

into

- x^{3} + 3 x^{2} - 6 \Rightarrow y = 14

Maximum (- 2, 14)

Plug the extreme point

x = 0

into

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

Minimum (0, - 6)

Plug the extreme point

x = 2

into

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

Maximum (2, - 2)

Plug the extreme point

x = 5

into

- x^{3} + 3 x^{2} - 6 \Rightarrow y = - 56

Minimum (5, - 56)

Maximum (- 2, 14), Minimum (0, - 6), Maximum (2, - 2), Minimum (5, - 56)

\frac{2 x - 2}{x ^{2} + x} + 1

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

f (x) = \frac{- 4}{x ^{2} - 9}

f (x, y) = e^{\frac{( x ^{2} + y ^{2} )}{2}}

f (x, y) = e^{3 x y}

What is the extreme f(x)=-x^3+3x^2-6,-2<= x<= 5 ?
The extreme f(x)=-x^3+3x^2-6,-2<= x<= 5 is Maximum(-2,14),Minimum(0,-6),Maximum(2,-2),Minimum(5,-56)