What is the extreme f(x)=2x^3-24x^2+72x+7,(3,7) ?

The extreme f(x)=2x^3-24x^2+72x+7,(3,7) is Minimum(6,7)

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extreme f(x)=2x^3-24x^2+72x+7,(3,7)

extreme points

f (x) = 2 x^{3} - 24 x^{2} + 72 x + 7, (3, 7)

Minimum (6, 7)

Solution steps

Find the critical points:

x = 6

Domain of

2 x^{3} - 24 x^{2} + 72 x + 7, (3, 7) : 3 < x < 7

Find intervals:

Decreasing

: 3 < x < 6,

Increasing

: 6 < x < 7

Plug the extreme point

x = 6

into

2 x^{3} - 24 x^{2} + 72 x + 7 \Rightarrow y = 7

Minimum (6, 7)

x^{4} + y^{4} - 4 x y + x

y = x^{3} + 3 x^{2} + 1

f (x) = - 4 x^{3} - 3 x^{2} + 6 x + 1

150 (x y)^{2} + (0.5 x + 2 y - 2)^{2}

3 x^{2} - 5 x - 1

What is the extreme f(x)=2x^3-24x^2+72x+7,(3,7) ?
The extreme f(x)=2x^3-24x^2+72x+7,(3,7) is Minimum(6,7)