What is the range of (sqrt(2+x))/(x-5) ?

The range of (sqrt(2+x))/(x-5) is -infinity <f(x)<infinity

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range of (sqrt(2+x))/(x-5)

range

\frac{2 + x}{x - 5}

- \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{2 + x}{x - 5} : - 2 \leq x < 5 or x > 5

Extreme Points of

\frac{2 + x}{x - 5} :

Maximum

(- 2, 0)

Find the range for the interval

- 2 \leq x < 5 : - \infty < f (x) \leq 0

Find the range for the interval

5 < 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

- 3 x - 9

f (x) = \frac{1}{3 x}

(5.13 e^{1.14 j})^{2}

8 x^{2} - x + 1

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

What is the range of (sqrt(2+x))/(x-5) ?
The range of (sqrt(2+x))/(x-5) is -infinity <f(x)<infinity