What is the extreme f(x)=4x^3-3x^2-6x+3,(0,10) ?

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

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extreme f(x)=4x^3-3x^2-6x+3,(0,10)

extreme points

f (x) = 4 x^{3} - 3 x^{2} - 6 x + 3, (0, 10)

Minimum (1, - 2)

Solution steps

Find the critical points:

x = 1

Domain of

4 x^{3} - 3 x^{2} - 6 x + 3, (0, 10) : 0 < x < 10

Find intervals:

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < 10

Plug the extreme point

x = 1

into

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

Minimum (1, - 2)

x^{2} y + x y^{2} + 3 x y

\frac{x ^{2} + 4 x + 3}{x ^{3} - 2 x ^{2} - 5 x + 6}

f (x, y) = 4 x^{3} + y^{3} - 12 x - 3 y

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

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

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