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 points

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

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)

x - 2

f (x) = \frac{( x + 5 )}{( x + 1 )}

f (x) = x^{2} + 14 x + 33

f (x) = x 1 - x^{2}

\frac{x ^{2} - 2 x - 15}{x ^{2} + 4 x}

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)