What is the domain of f(x)=(sqrt(3-x))/(x^2-6x+8) ?

The domain of f(x)=(sqrt(3-x))/(x^2-6x+8) is x<2\lor 2<x<= 3

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domain of f(x)=(sqrt(3-x))/(x^2-6x+8)

domain

f (x) = \frac{3 - x}{x ^{2} - 6 x + 8}

x < 2 or 2 < x \leq 3

Interval Notation

(- \infty, 2) \cup (2, 3]

Solution steps

Find non-negative values for radicals:

x \leq 3

Find undefined (singularity) points:

x = 4, x = 2

Combine real regions and undefined points for final function domain

x < 2 or 2 < x \leq 3

f (x) = 3 x^{3} + 8

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

y = - 4 cos (2 x)

f (x) = - \frac{1}{2} x^{2} + 5 x - 2

f (x) = \frac{x + 8}{x - 5}

What is the domain of f(x)=(sqrt(3-x))/(x^2-6x+8) ?
The domain of f(x)=(sqrt(3-x))/(x^2-6x+8) is x<2\lor 2<x<= 3