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

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

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

range

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

- \infty < f (x) < \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

f (x) = 2 x - 3

f (x) = - 0.1 (x - 4)^{2} + 4.2

2 (1)^{4} - 11 (1)^{2}

f (x) = \frac{( x + 3 )}{( x - 8 )}

f (x) = 5 x - 4 y = 20

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