What is the domain of f(x)=(sqrt(2x^2+1))/(3x-5) ?

The domain of f(x)=(sqrt(2x^2+1))/(3x-5) is x 5/3

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domain of f(x)=(sqrt(2x^2+1))/(3x-5)

domain

f (x) = \frac{2 x ^{2} + 1}{3 x - 5}

x < \frac{5}{3} or x > \frac{5}{3}

Interval Notation

(- \infty, \frac{5}{3}) \cup (\frac{5}{3}, \infty)

Solution steps

Find non-negative values for radicals:

- \infty < x < \infty

Find undefined (singularity) points:

x = \frac{5}{3}

Combine real regions and undefined points for final function domain

x < \frac{5}{3} or x > \frac{5}{3}

f (x) = \frac{x ^{2} + 2 x - 3}{( x - 1 ) ( x ^{2} - 4 )}

f (x) = - 3 + x + 1

\frac{3 x ^{2}}{x ^{2} - 16}

f (x) = x x^{2} - 2

f (x) = - lo g_{3} (x)

What is the domain of f(x)=(sqrt(2x^2+1))/(3x-5) ?
The domain of f(x)=(sqrt(2x^2+1))/(3x-5) is x< 5/3 \lor x> 5/3