What is the critical (6x^2+2x-3)/(x-4) ?

The critical (6x^2+2x-3)/(x-4) is x=(24-sqrt(606))/6 ,x=(24+sqrt(606))/6

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critical (6x^2+2x-3)/(x-4)

critical points

\frac{6 x ^{2} + 2 x - 3}{x - 4}

x = \frac{24 - 606}{6}, x = \frac{24 + 606}{6}

Solution steps

\frac{6 x ^{2} + 2 x - 3}{x - 4}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{24 - 606}{6}, x = \frac{24 + 606}{6}, x = 4

Identify critical points not in

f (x)

domain

Domain of

\frac{6 x ^{2} + 2 x - 3}{x - 4} : x < 4 or x > 4

\frac{6 x ^{2} + 2 x - 3}{x - 4}

is not defined at

x = 4

, therefore:

x = \frac{24 - 606}{6}, x = \frac{24 + 606}{6}

f (x) = 2 x^{3} + 3 x^{2} - 72 x + 7

x e^{6 x^{2}}

f (x) = x (x - 6)

f (x) = x^{6} e^{x} - 8

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

What is the critical (6x^2+2x-3)/(x-4) ?
The critical (6x^2+2x-3)/(x-4) is x=(24-sqrt(606))/6 ,x=(24+sqrt(606))/6