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

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

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critical f(x)= x/(x^2-4x+3)

critical points

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

x = - 3, x = 3

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

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

Identify critical points not in

f (x)

domain

Domain of

\frac{x}{x ^{2} - 4 x + 3} : x < 1 or 1 < x < 3 or x > 3

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

is not defined at

x = 1, x = 3

, therefore:

x = - 3, x = 3

h (x) = \frac{4 x}{7 x - 3}

5 - 8 e^{x}

y = 9 sin (14 x)

- x + 3

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

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