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

The domain of f(x)=sqrt((6-3x)/(x^2+3x-2)) is x<(-3-sqrt(17))/2 \lor (-3+sqrt(17))/2 <x<= 2

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

domain

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

x < \frac{- 3 - 17}{2} or \frac{- 3 + 17}{2} < x \leq 2

Interval Notation

(- \infty, \frac{- 3 - 17}{2}) \cup (\frac{- 3 + 17}{2}, 2]

Solution steps

Find non-negative values for radicals:

x < \frac{- 3 - 17}{2} or \frac{17 - 3}{2} < x \leq 2

Find undefined (singularity) points:

x = \frac{- 3 + 17}{2}, x = \frac{- 3 - 17}{2}

Combine real regions and undefined points for final function domain

x < \frac{- 3 - 17}{2} or \frac{- 3 + 17}{2} < x \leq 2

f^{8}

f^{9}

f (x) = \frac{6}{x + 4}

\frac{x}{7 x - 6}

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

What is the domain of f(x)=sqrt((6-3x)/(x^2+3x-2)) ?
The domain of f(x)=sqrt((6-3x)/(x^2+3x-2)) is x<(-3-sqrt(17))/2 \lor (-3+sqrt(17))/2 <x<= 2