What is the range of (sqrt(x-2))/(x-9) ?

The range of (sqrt(x-2))/(x-9) is -infinity <f(x)<infinity

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range of (sqrt(x-2))/(x-9)

range

\frac{x - 2}{x - 9}

- \infty < f (x) < \infty

Interval Notation

(- \infty, \infty)

Solution steps

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

Domain of

\frac{x - 2}{x - 9} : 2 \leq x < 9 or x > 9

Extreme Points of

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

Maximum

(2, 0)

Find the range for the interval

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

Find the range for the interval

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

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

f (x) \leq 0 or f (x) > 0

- \infty < f (x) < \infty

f (x) = 1 + 4 - y^{2}

x^{3} - 27 x + 9

f (x) = - 3 x^{4} + 30 x^{2} - 27

f (x) = \frac{4}{2 x - 1}

y - x = 5

What is the range of (sqrt(x-2))/(x-9) ?
The range of (sqrt(x-2))/(x-9) is -infinity <f(x)<infinity