What is the critical 4x^3+48x^2+6x+3 ?

The critical 4x^3+48x^2+6x+3 is x=-(8+sqrt(62))/2 ,x=(-8+sqrt(62))/2

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critical 4x^3+48x^2+6x+3

critical points

4 x^{3} + 48 x^{2} + 6 x + 3

x = - \frac{8 + 62}{2}, x = \frac{- 8 + 62}{2}

Solution steps

4 x^{3} + 48 x^{2} + 6 x + 3

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{8 + 62}{2}, x = \frac{- 8 + 62}{2}

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = - \frac{8 + 62}{2}, x = \frac{- 8 + 62}{2}

f (x) = x^{2} - 12 x + 36

f (x) = \frac{x ^{2} + 9 x + 20}{4 x + 16}

x^{6}

y = 3 x, (2, 6)

x - 3 y = 6

What is the critical 4x^3+48x^2+6x+3 ?
The critical 4x^3+48x^2+6x+3 is x=-(8+sqrt(62))/2 ,x=(-8+sqrt(62))/2