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

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### 2 Answers

The range of a function represents the y-values of a graph. The lowest value in the domain of f(x) is x>1, since the square root of a negative is undefined and the denominator can't equal 0. Using a graphing calculator, the minimum value for f(x) is 2, which happens at x=2. The graph rises to infinity. The range of f(x) is f(x)>=2, or 2<=f(x)<infinity.

The range of a function is the set of the values of f(x) when x lies in the domain.

The domain of f(x) is all values where x – 1 > 0 because if the denominator is 0, we get an indeterminate number and for x - 1 < 0, the square root is a complex number.

=> x > 1

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

f'(x) = [1*sqrt(x - 1) - x*(1/2)(1/sqrt(x - 1))]/(x - 1)

=> f'(x) = [(x - 1) - x/2]/(x - 1)^(3/2)

=> f'(x) = (x - 2)/2*(x - 1)^(3/2)

equating f'(x) = 0, we get x = 2

The function has the minimum value when x = 2

f(2) = 2/sqrt 1 = 2

If x > 1, f(x) >= 2

**The range of the function is (inf., 2]**