What is the range of f(x)= 4/(sqrt(x^2-9)) ?

The range of f(x)= 4/(sqrt(x^2-9)) is f(x)>0

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range of f(x)= 4/(sqrt(x^2-9))

range

f (x) = \frac{4}{x ^{2} - 9}

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{4}{x ^{2} - 9} : x < - 3 or x > 3

Extreme Points of

\frac{4}{x ^{2} - 9} :

None

Find the range for the interval

- \infty < x < - 3 : 0 < f (x) < \infty

Find the range for the interval

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

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

f (x) > 0 or f (x) > 0

f (x) > 0

3

f (x) = \frac{x ^{2} + 9}{9 x ^{2} - 26 x - 3}

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

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

f (x) = 5 - 4 x

What is the range of f(x)= 4/(sqrt(x^2-9)) ?
The range of f(x)= 4/(sqrt(x^2-9)) is f(x)>0