What is the range of f(x)= 1/(sqrt(x-2)) ?

The range of f(x)= 1/(sqrt(x-2)) is f(x)>0

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range of f(x)= 1/(sqrt(x-2))

range

f (x) = \frac{1}{x - 2}

f (x) > 0

Interval Notation

(0, \infty)

Solution steps

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

Domain of

\frac{1}{x - 2} : x > 2

Extreme Points of

\frac{1}{x - 2} :

None

Find the range for the interval

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

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

f (x) > 0

x^{2} - 25

f (x) = \frac{x - 1}{ln ( x ^{2} )}

f (x) = lo g_{a} (x)

2 x^{3} - 24 x - 6

f (x) = \frac{x}{2} + 1

What is the range of f(x)= 1/(sqrt(x-2)) ?
The range of f(x)= 1/(sqrt(x-2)) is f(x)>0