What is the range of 1/(x^2-9) ?

The range of 1/(x^2-9) is f(x) 0

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range of 1/(x^2-9)

range

\frac{1}{x ^{2} - 9}

f (x) \leq - \frac{1}{9} or f (x) > 0

Interval Notation

(- \infty, - \frac{1}{9}] \cup (0, \infty)

Solution steps

Rewrite as

\frac{1}{x ^{2} - 9} = y

Multiply both sides by

x^{2} - 9

1 = y (x^{2} - 9)

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

Discriminant

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

36 y^{2} + 4 y \geq 0 : y \leq - \frac{1}{9} or y \geq 0

Check if the range interval endpoints are included

y = - \frac{1}{9} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{1}{9} or f (x) > 0

f (x) = 1 + \frac{1}{2} x

\frac{1}{7 x ^{3} - 5 x}

\frac{( x + 10 )}{x ^{2} - 100}

f (x) = \frac{2 x ^{2} - 3 x - 2}{x ^{2} - 4}

x - 2 y = 12, (- 8, - 7)

What is the range of 1/(x^2-9) ?
The range of 1/(x^2-9) is f(x)<=-1/9 \lor f(x)>0