What is the extreme 2x^3+3x^2-336x ?

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

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extreme 2x^3+3x^2-336x

extreme points

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)

f (x) = \frac{6 x}{( x ^{2} - 4 )}

f (x) = \frac{x ^{3}}{6 x ^{2} + 4}

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

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

x^{2} + 2 y^{2} - 2 x - 12 y + 19

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