What is the critical f(x)=(x-3)/(x^2-5x+6) ?

The critical f(x)=(x-3)/(x^2-5x+6) is None

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critical f(x)=(x-3)/(x^2-5x+6)

critical points

f (x) = \frac{x - 3}{x ^{2} - 5 x + 6}

None

Solution steps

\frac{x - 3}{x ^{2} - 5 x + 6}

Find where

f^{'} (x)

is equal to zero or undefined

x = 2

Identify critical points not in

f (x)

domain

Domain of

\frac{x - 3}{x ^{2} - 5 x + 6} : x < 2 or 2 < x < 3 or x > 3

\frac{x - 3}{x ^{2} - 5 x + 6}

is not defined at

x = 2

, therefore:

None

\frac{1}{2} x^{4} - 3 x y + 3 y^{4}

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

f (x) = ∣ x ∣ + \frac{x}{7}

f (x) = 18 cos (x) + 9 sin^{2} (x)

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

What is the critical f(x)=(x-3)/(x^2-5x+6) ?
The critical f(x)=(x-3)/(x^2-5x+6) is None