What is the extreme f(x)=2x^3+3x^2-336x ?

The extreme f(x)=2x^3+3x^2-336x is Maximum(-8,1856),Minimum(7,-1519)

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extreme f(x)=2x^3+3x^2-336x

extreme points

f (x) = 2 x^{3} + 3 x^{2} - 336 x

Maximum (- 8, 1856), Minimum (7, - 1519)

Solution steps

f^{'} (x) = 6 x^{2} + 6 x - 336

Find intervals:

Increasing

: - \infty < x < - 8,

Decreasing

: - 8 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = - 8

into

2 x^{3} + 3 x^{2} - 336 x : 1856

Maximum (- 8, 1856)

Plug

x = 7

into

2 x^{3} + 3 x^{2} - 336 x : - 1519

Minimum (7, - 1519)

Maximum (- 8, 1856), Minimum (7, - 1519)

y = x^{4} + 6 x^{3} + 12 x^{2} + 8 x + 10

f (x, y) = x^{2} - 2 x y + 3 y^{2}

f (x) = \frac{e ^{x}}{1 + x ^{2}}

f (x, y) = 6 x y - 2 x^{2} y - 3 x y^{2}

f (x, y) = x^{2} + 3 x y

What is the extreme f(x)=2x^3+3x^2-336x ?
The extreme f(x)=2x^3+3x^2-336x is Maximum(-8,1856),Minimum(7,-1519)