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

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

6 - x

f (x) = \frac{( 3 x + 2 )}{x ^{2} - 7 x}

f (x) = \frac{5}{2 5 x + 6}

f (x) = 2 x^{2} + 10 x + 1

3 x - 2 y = 8

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