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

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

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

domain

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

x < \frac{3}{2}

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \leq \frac{3}{2}

Find undefined (singularity) points:

x = \frac{3}{2}

Combine real regions and undefined points for final function domain

x < \frac{3}{2}

\frac{x + 1}{x}

f (x) = ∣ x ∣ + x^{2}

y = 3 x, (1, 3)

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

\frac{( x + 6 ) ( x + 4 )}{x ( x - 5 ) ( x + 2 )}

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