What is the range of f(x)=(sqrt(x+3))/x ?

The range of f(x)=(sqrt(x+3))/x is -infinity <f(x)<infinity

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range of f(x)=(sqrt(x+3))/x

range

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

- \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 + 3}{x} : - 3 \leq x < 0 or x > 0

Extreme Points of

\frac{x + 3}{x} :

Maximum

(- 3, 0)

Find the range for the interval

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

Find the range for the interval

0 < 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

x^{3} + 5

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

f (x) = x^{2} + 3 x + 2

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

\frac{1}{2 x - 6}

What is the range of f(x)=(sqrt(x+3))/x ?
The range of f(x)=(sqrt(x+3))/x is -infinity <f(x)<infinity