What is the extreme f(x)=3x^3-3x^2+12x-5,-1<= x<= 1 ?

The extreme f(x)=3x^3-3x^2+12x-5,-1<= x<= 1 is Minimum(-1,-23),Maximum(1,7)

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extreme f(x)=3x^3-3x^2+12x-5,-1<= x<= 1

extreme points

f (x) = 3 x^{3} - 3 x^{2} + 12 x - 5, - 1 \leq x \leq 1

Minimum (- 1, - 23), Maximum (1, 7)

Solution steps

Find the critical points:

x = - 1, x = 1

Domain of

3 x^{3} - 3 x^{2} + 12 x - 5, - 1 \leq x \leq 1 : - 1 \leq x \leq 1

Find intervals:

Increasing

: - 1 < x < 1

Plug the extreme point

x = - 1

into

3 x^{3} - 3 x^{2} + 12 x - 5 \Rightarrow y = - 23

Minimum (- 1, - 23)

Plug the extreme point

x = 1

into

3 x^{3} - 3 x^{2} + 12 x - 5 \Rightarrow y = 7

Maximum (1, 7)

Minimum (- 1, - 23), Maximum (1, 7)

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

f (x) = - 2 x^{3} + 21 x^{2} - 36 x + 3

f (x) = 3 x^{2} + 5 y^{2} - 10 x y - 6 y + 1

9600 + 280 x - 10 x^{2}

What is the extreme f(x)=3x^3-3x^2+12x-5,-1<= x<= 1 ?
The extreme f(x)=3x^3-3x^2+12x-5,-1<= x<= 1 is Minimum(-1,-23),Maximum(1,7)