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

The extreme f(x)=2x^3+3x^2-36x+15 is Maximum(-3,96),Minimum(2,-29)

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extreme f(x)=2x^3+3x^2-36x+15

extreme points

f (x) = 2 x^{3} + 3 x^{2} - 36 x + 15

Maximum (- 3, 96), Minimum (2, - 29)

Solution steps

f^{'} (x) = 6 x^{2} + 6 x - 36

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 3

into

2 x^{3} + 3 x^{2} - 36 x + 15 : 96

Maximum (- 3, 96)

Plug

x = 2

into

2 x^{3} + 3 x^{2} - 36 x + 15 : - 29

Minimum (2, - 29)

Maximum (- 3, 96), Minimum (2, - 29)

y = e^{(2 - x)} + x^{2} - \frac{x}{2}

cos (2 x)

f (x) = - 6 x^{2} - 9 x y - 7 y^{2} - 99 x - 96 y + 5

f (x) = lo g_{e} (x)

P (x, y) = x + y^{2} - 10 x^{2} y^{2} + 9 y^{6}

What is the extreme f(x)=2x^3+3x^2-36x+15 ?
The extreme f(x)=2x^3+3x^2-36x+15 is Maximum(-3,96),Minimum(2,-29)