What is the extreme f(x)=2x^3-6x^2-48x+10 ?

The extreme f(x)=2x^3-6x^2-48x+10 is Maximum(-2,66),Minimum(4,-150)

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extreme f(x)=2x^3-6x^2-48x+10

extreme points

f (x) = 2 x^{3} - 6 x^{2} - 48 x + 10

Maximum (- 2, 66), Minimum (4, - 150)

Solution steps

f^{'} (x) = 6 x^{2} - 12 x - 48

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 2

into

2 x^{3} - 6 x^{2} - 48 x + 10 : 66

Maximum (- 2, 66)

Plug

x = 4

into

2 x^{3} - 6 x^{2} - 48 x + 10 : - 150

Minimum (4, - 150)

Maximum (- 2, 66), Minimum (4, - 150)

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f (x) = x^{2} x + 5

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

f (x) = x^{2} - 8 x + 4

What is the extreme f(x)=2x^3-6x^2-48x+10 ?
The extreme f(x)=2x^3-6x^2-48x+10 is Maximum(-2,66),Minimum(4,-150)