What is the domain of f(x)= 1/(sqrt(2x^2-7x+3)) ?

The domain of f(x)= 1/(sqrt(2x^2-7x+3)) is x 3

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domain of f(x)= 1/(sqrt(2x^2-7x+3))

domain

f (x) = \frac{1}{2 x ^{2} - 7 x + 3}

x < \frac{1}{2} or x > 3

Interval Notation

(- \infty, \frac{1}{2}) \cup (3, \infty)

Solution steps

Find non-negative values for radicals:

x \leq \frac{1}{2} or x \geq 3

Find undefined (singularity) points:

x = 3, x = \frac{1}{2}

Combine real regions and undefined points for final function domain

x < \frac{1}{2} or x > 3

f (x) = 4 x^{6} - 6 x^{4}

(0, 6), (2, 1)

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

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

\frac{x + 4}{x - 4}

What is the domain of f(x)= 1/(sqrt(2x^2-7x+3)) ?
The domain of f(x)= 1/(sqrt(2x^2-7x+3)) is x< 1/2 \lor x>3