What is the range of (x+4)/(x^2-9) ?

The range of (x+4)/(x^2-9) is f(x) = (sqrt(7))/(18)-2/9

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range of (x+4)/(x^2-9)

range

\frac{x + 4}{x ^{2} - 9}

f (x) \leq - \frac{7}{18} - \frac{2}{9} or f (x) \geq \frac{7}{18} - \frac{2}{9}

Interval Notation

(- \infty, - \frac{7}{18} - \frac{2}{9}] \cup [\frac{7}{18} - \frac{2}{9}, \infty)

Solution steps

Rewrite as

\frac{x + 4}{x ^{2} - 9} = y

Multiply both sides by

x^{2} - 9

x + 4 = y (x^{2} - 9)

The range is the set of y for which the discriminant is greater or equal to zero

Discriminant

x + 4 = y (x^{2} - 9) : 1 + 36 y^{2} + 16 y

1 + 36 y^{2} + 16 y \geq 0 : y \leq - \frac{7}{18} - \frac{2}{9} or y \geq \frac{7}{18} - \frac{2}{9}

Check if the range interval endpoints are included

y = - \frac{7}{18} - \frac{2}{9} :

Included

y = \frac{7}{18} - \frac{2}{9} :

Included

Therefore the range is

f (x) \leq - \frac{7}{18} - \frac{2}{9} or f (x) \geq \frac{7}{18} - \frac{2}{9}

f (x) = (x - 1)^{\frac{4}{3}}

f (x) = 5 + x + 8

\frac{6}{6 - x}

f (x) = \frac{4 x ^{2} + 5}{6 x + 4}

What is the range of (x+4)/(x^2-9) ?
The range of (x+4)/(x^2-9) is f(x)<=-(sqrt(7))/(18)-2/9 \lor f(x)>= (sqrt(7))/(18)-2/9