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

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

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

domain

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

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

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \geq 1

Find undefined (singularity) points:

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

Combine real regions and undefined points for final function domain

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

f (x) = \frac{x - 7}{x + 4}

f (x) = \frac{1}{x - 9}

- 6 cos (5 x)

6 x^{2} + 8 x - 1

f (x) = \frac{10 x - 1}{3 - 5 x}

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