What is the domain of f(x)=(sqrt(x))/(x^2-16x+48) ?

The domain of f(x)=(sqrt(x))/(x^2-16x+48) is 0 12

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domain of f(x)=(sqrt(x))/(x^2-16x+48)

domain

f (x) = \frac{x}{x ^{2} - 16 x + 48}

0 \leq x < 4 or 4 < x < 12 or x > 12

Interval Notation

[0, 4) \cup (4, 12) \cup (12, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = 12, x = 4

Combine real regions and undefined points for final function domain

0 \leq x < 4 or 4 < x < 12 or x > 12

- \frac{25}{( 6 + t ) ^{2}}

f (x) = \frac{( - 9 - 7 x )}{3}

\frac{5 x - 15}{2 x - 9}

f (x) = \frac{3}{5} x^{3} - 7 x + 4

f (x) = \frac{x + 6}{x - 5}

What is the domain of f(x)=(sqrt(x))/(x^2-16x+48) ?
The domain of f(x)=(sqrt(x))/(x^2-16x+48) is 0<= x<4\lor 4<x<12\lor x>12