What is the extreme f(x)=2x^3-96x-3 ?

The extreme f(x)=2x^3-96x-3 is Maximum(-4,253),Minimum(4,-259)

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extreme f(x)=2x^3-96x-3

extreme points

f (x) = 2 x^{3} - 96 x - 3

Maximum (- 4, 253), Minimum (4, - 259)

Solution steps

f^{'} (x) = 6 x^{2} - 96

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 4

into

2 x^{3} - 96 x - 3 : 253

Maximum (- 4, 253)

Plug

x = 4

into

2 x^{3} - 96 x - 3 : - 259

Minimum (4, - 259)

Maximum (- 4, 253), Minimum (4, - 259)

x ln (\frac{x}{9}) - x

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

f (x, y) = x^{2} - 2 x + y^{2} + 1

\frac{x + 3}{x - 2}

\frac{3}{4} \frac{2}{5}

What is the extreme f(x)=2x^3-96x-3 ?
The extreme f(x)=2x^3-96x-3 is Maximum(-4,253),Minimum(4,-259)