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

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

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domain of f(x)= x/(2x^2-5)

domain

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

x < - \frac{5}{2} or - \frac{5}{2} < x < \frac{5}{2} or x > \frac{5}{2}

Interval Notation

(- \infty, - \frac{5}{2}) \cup (- \frac{5}{2}, \frac{5}{2}) \cup (\frac{5}{2}, \infty)

Solution steps

Find undefined (singularity) points:

x = \frac{5}{2}, x = - \frac{5}{2}

The function domain

x < - \frac{5}{2} or - \frac{5}{2} < x < \frac{5}{2} or x > \frac{5}{2}

f (x) = ln (2 - x)

6 - x

\frac{2 x + 5}{3 x + 1}

f (x) = x^{3} + 2 x^{2} + x + 10

f (x) = x^{6} - 5 x^{4} - 6 x^{2}

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