What is the critical x^3+6x^2-13x+42 ?

The critical x^3+6x^2-13x+42 is x=-(6+5sqrt(3))/3 ,x=(-6+5sqrt(3))/3

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critical x^3+6x^2-13x+42

critical points

x^{3} + 6 x^{2} - 13 x + 42

x = - \frac{6 + 5 3}{3}, x = \frac{- 6 + 5 3}{3}

Solution steps

x^{3} + 6 x^{2} - 13 x + 42

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{6 + 5 3}{3}, x = \frac{- 6 + 5 3}{3}

Identify critical points not in

f (x)

domain

Domain of

x^{3} + 6 x^{2} - 13 x + 42 : - \infty < x < \infty

All critical points are in domain

x = - \frac{6 + 5 3}{3}, x = \frac{- 6 + 5 3}{3}

5 x^{2} ln (x)

f (x) = I n^{2} 2 x

f (x) = - 20 - 40 x - 25 x^{2}

x^{3} - 6 x^{2} - 15 x + 40

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

What is the critical x^3+6x^2-13x+42 ?
The critical x^3+6x^2-13x+42 is x=-(6+5sqrt(3))/3 ,x=(-6+5sqrt(3))/3