What is the extreme f(x)=2x^3+15x^2-36x+7,-6<x<2 ?

The extreme f(x)=2x^3+15x^2-36x+7,-6<x<2 is Minimum(1,-12)

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extreme f(x)=2x^3+15x^2-36x+7,-6<x<2

extreme points

f (x) = 2 x^{3} + 15 x^{2} - 36 x + 7, - 6 < x < 2

Minimum (1, - 12)

Solution steps

Find the critical points:

x = 1

Domain of

2 x^{3} + 15 x^{2} - 36 x + 7, - 6 < x < 2 : - 6 < x < 2

Find intervals:

Decreasing

: - 6 < x < 1,

Increasing

: 1 < x < 2

Plug the extreme point

x = 1

into

2 x^{3} + 15 x^{2} - 36 x + 7 \Rightarrow y = - 12

Minimum (1, - 12)

f (x) = 13 e^{- x}, (0, 4)

- y^{2} + 120 y

f (x) = x - \frac{1}{x} + \frac{2}{x ^{3}}, [0, \infty]

y = 2 x_{1} + 4 x_{2}

f (x, y) = - 25 - x^{2} - y^{2}

What is the extreme f(x)=2x^3+15x^2-36x+7,-6<x<2 ?
The extreme f(x)=2x^3+15x^2-36x+7,-6<x<2 is Minimum(1,-12)