What is the domain of f(x)=sqrt(6+(40)/x) ?

The domain of f(x)=sqrt(6+(40)/x) is x 0

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domain of f(x)=sqrt(6+(40)/x)

domain

f (x) = 6 + \frac{40}{x}

x \leq - \frac{20}{3} or x > 0

Interval Notation

(- \infty, - \frac{20}{3}] \cup (0, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - \frac{20}{3} or x > 0

Find undefined (singularity) points:

x = 0

Combine real regions and undefined points for final function domain

x \leq - \frac{20}{3} or x > 0

f (x) = - 5 x^{2} - 40 x - 77

y = \frac{( - x - 2 )}{( 5 x + 3 )}

x + 13 y = - 8

\frac{1}{x + 8} + ln (\frac{1}{x} - \frac{1}{1 - x})

f (x) = 9 - x^{2}

What is the domain of f(x)=sqrt(6+(40)/x) ?
The domain of f(x)=sqrt(6+(40)/x) is x<=-20/3 \lor x>0