What is the domain of f(x)=(sqrt(4x+7))/(9-x^2) ?

The domain of f(x)=(sqrt(4x+7))/(9-x^2) is -7/4 3

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domain of f(x)=(sqrt(4x+7))/(9-x^2)

domain

f (x) = \frac{4 x + 7}{9 - x ^{2}}

- \frac{7}{4} \leq x < 3 or x > 3

Interval Notation

[- \frac{7}{4}, 3) \cup (3, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - \frac{7}{4}

Find undefined (singularity) points:

x = 3, x = - 3

Combine real regions and undefined points for final function domain

- \frac{7}{4} \leq x < 3 or x > 3

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

f (x) = lo g_{10} (x + 4)

f (x) = \frac{x + 7}{24 - x ^{2} - 49}

f (x) = \frac{7 - 3 x}{x - 2}

f (x) = x + 3 \cdot (\frac{8}{x})

What is the domain of f(x)=(sqrt(4x+7))/(9-x^2) ?
The domain of f(x)=(sqrt(4x+7))/(9-x^2) is -7/4 <= x<3\lor x>3