What is the range of f(x)=sqrt(x/(x-1)) ?

The range of f(x)=sqrt(x/(x-1)) is 0 1

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range of f(x)=sqrt(x/(x-1))

range

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

0 \leq f (x) < 1 or f (x) > 1

Interval Notation

[0, 1) \cup (1, \infty)

Solution steps

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

Domain of

\frac{x}{x - 1} : x \leq 0 or x > 1

Extreme Points of

\frac{x}{x - 1} :

None

Find the range for the interval

- \infty < x \leq 0 : 0 \leq f (x) < 1

Find the range for the interval

1 < x < \infty : 1 < f (x) < \infty

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

0 \leq f (x) < 1 or f (x) > 1

0 \leq f (x) < 1 or f (x) > 1

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

f (x) = 3 x^{5} - 5 x^{3}

f (x) = x^{5} + 3 x

f (x) = cos (9 x)

- \frac{1}{x ^{4}} - 3

What is the range of f(x)=sqrt(x/(x-1)) ?
The range of f(x)=sqrt(x/(x-1)) is 0<= f(x)<1\lor f(x)>1