What is the range of (sqrt(2+x))/(3-x) ?

The range of (sqrt(2+x))/(3-x) is -infinity <f(x)<infinity

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range of (sqrt(2+x))/(3-x)

range

\frac{2 + 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{2 + x}{3 - x} : - 2 \leq x < 3 or x > 3

Extreme Points of

\frac{2 + x}{3 - x} :

Minimum

(- 2, 0)

Find the range for the interval

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

Find the range for the interval

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

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

f (x) \geq 0 or f (x) < 0

- \infty < f (x) < \infty

6 x - 3 y = 18

f (x) = \frac{x ^{5}}{7}

f (x) = 500 + 10 x^{2}

f (x) = x^{5} - 5 x^{3}

x^{2} + 4

What is the range of (sqrt(2+x))/(3-x) ?
The range of (sqrt(2+x))/(3-x) is -infinity <f(x)<infinity