What is the domain of f(x)=-1/(sqrt(8x-3-4x^2)) ?

The domain of f(x)=-1/(sqrt(8x-3-4x^2)) is 1/2 <x< 3/2

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domain of f(x)=-1/(sqrt(8x-3-4x^2))

domain

f (x) = - \frac{1}{8 x - 3 - 4 x ^{2}}

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

Interval Notation

(\frac{1}{2}, \frac{3}{2})

Solution steps

Find non-negative values for radicals:

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

Find undefined (singularity) points:

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

Combine real regions and undefined points for final function domain

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

arccos (ln (3 x))

f (x) = \frac{8}{x + 7}

f (x) = \frac{3}{x ^{2} - 36}

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

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

What is the domain of f(x)=-1/(sqrt(8x-3-4x^2)) ?
The domain of f(x)=-1/(sqrt(8x-3-4x^2)) is 1/2 <x< 3/2