What is the extreme f(x)=4x^3-6x^2,-1<= x<= 1 ?

The extreme f(x)=4x^3-6x^2,-1<= x<= 1 is Minimum(-1,-10),Maximum(0,0),Minimum(1,-2)

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extreme f(x)=4x^3-6x^2,-1<= x<= 1

extreme points

f (x) = 4 x^{3} - 6 x^{2}, - 1 \leq x \leq 1

Minimum (- 1, - 10), Maximum (0, 0), Minimum (1, - 2)

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1

Plug the extreme point

x = - 1

into

4 x^{3} - 6 x^{2} \Rightarrow y = - 10

Minimum (- 1, - 10)

Plug the extreme point

x = 0

into

4 x^{3} - 6 x^{2} \Rightarrow y = 0

Maximum (0, 0)

Plug the extreme point

x = 1

into

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

Minimum (1, - 2)

Minimum (- 1, - 10), Maximum (0, 0), Minimum (1, - 2)

f (x) = \frac{10 x}{3 x ^{2} + 3}

f (x) = (x - 4)^{2} + (y - 4)^{2} - \frac{4}{5}

3 x^{4} + 7 x^{3} - 1 8^{2} - 32

f (x) = 0.2 x^{2} - 20 x + 900

14 + 4 x - x^{2}

What is the extreme f(x)=4x^3-6x^2,-1<= x<= 1 ?
The extreme f(x)=4x^3-6x^2,-1<= x<= 1 is Minimum(-1,-10),Maximum(0,0),Minimum(1,-2)