What is the critical f(x)=(x^2+3)/(x-1) ?

The critical f(x)=(x^2+3)/(x-1) is x=-1,x=3

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critical f(x)=(x^2+3)/(x-1)

critical points

f (x) = \frac{x ^{2} + 3}{x - 1}

x = - 1, x = 3

Solution steps

\frac{x ^{2} + 3}{x - 1}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 3, x = 1

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2} + 3}{x - 1} : x < 1 or x > 1

\frac{x ^{2} + 3}{x - 1}

is not defined at

x = 1

, therefore:

x = - 1, x = 3

f (x) = 3 x^{4} + 4 x^{3} + 6 x^{2} - 4

f (x, y) = 2 x^{3} y - 29 x^{2} - 16 y

y = x^{2} + 2 x + 25

y = I n (x)

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

What is the critical f(x)=(x^2+3)/(x-1) ?
The critical f(x)=(x^2+3)/(x-1) is x=-1,x=3