What is the range of y=sqrt(7/(x-5)) ?

The range of y=sqrt(7/(x-5)) is f(x)>0

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range of y=sqrt(7/(x-5))

range

y = \frac{7}{x - 5}

f (x) > 0

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

\frac{7}{x - 5} : x > 5

Extreme Points of

\frac{7}{x - 5} :

None

Find the range for the interval

5 < x < \infty : 0 < f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

f (x) > 0

10 x + 25 y = 225

f (x) = x^{3} + x^{2} - 9 x - 9

f (x) = \frac{4}{x}

f (x) = (4 + 3 x)^{7}

f (x) = \frac{x + 4}{x ^{2} - 8 x + 16}