What is the critical 2x^3-6x^2-48x+7 ?

The critical 2x^3-6x^2-48x+7 is x=-2,x=4

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critical 2x^3-6x^2-48x+7

critical points

2 x^{3} - 6 x^{2} - 48 x + 7

x = - 2, x = 4

Solution steps

2 x^{3} - 6 x^{2} - 48 x + 7

Find where

f^{'} (x)

is equal to zero or undefined

x = - 2, x = 4

Identify critical points not in

f (x)

domain

Domain of

2 x^{3} - 6 x^{2} - 48 x + 7 : - \infty < x < \infty

All critical points are in domain

x = - 2, x = 4

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

2 x^{2} - 8 x + 6

f (x, y) = x^{2} + 2 x y + 2 y^{2} - 8 x + 6 y

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

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