What is the range of sqrt(((9+x))/(9-x)) ?

The range of sqrt(((9+x))/(9-x)) is f(x)>= 0

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range of sqrt(((9+x))/(9-x))

range

\frac{( 9 + x )}{9 - x}

f (x) \geq 0

Interval Notation

[0, \infty)

Solution steps

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

Domain of

\frac{9 + x}{9 - x} : - 9 \leq x < 9

Extreme Points of

\frac{9 + x}{9 - x} :

Minimum

(- 9, 0)

Find the range for the interval

- 9 \leq x < 9 : 0 \leq f (x) < \infty

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

f (x) \geq 0

f (x) = x + 1

f (x) = \frac{x - 5}{x + 1}

f (x) = x^{2} - 5 x

x^{2} - 18 x + 73

x + y = 0

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