What is the extreme f(x)=2x^3-96x+42 ?

The extreme f(x)=2x^3-96x+42 is Maximum(-4,298),Minimum(4,-214)

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extreme f(x)=2x^3-96x+42

extreme

Maximum (- 4, 298), Minimum (4, - 214)

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 + 42 : 298

Maximum (- 4, 298)

Plug

x = 4

into

2 x^{3} - 96 x + 42 : - 214

Minimum (4, - 214)

Maximum (- 4, 298), Minimum (4, - 214)

What is the extreme f(x)=2x^3-96x+42 ?
The extreme f(x)=2x^3-96x+42 is Maximum(-4,298),Minimum(4,-214)