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

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

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

extreme points

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

Maximum (- 1, 4), Minimum (0, 1), Maximum (1, 2)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 1

Domain of

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

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1

Plug the extreme point

x = - 1

into

- x^{3} + 2 x^{2} + 1 \Rightarrow y = 4

Maximum (- 1, 4)

Plug the extreme point

x = 0

into

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

Minimum (0, 1)

Plug the extreme point

x = 1

into

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

Maximum (1, 2)

Maximum (- 1, 4), Minimum (0, 1), Maximum (1, 2)

f (x) = \frac{1}{4} x^{2} + x^{3} - 2

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

f (x) = x^{3} - x^{2} - x + 4

f (x) = 2 x + \frac{8}{x}

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

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