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

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

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range of f(x)= 1/(x^2-2)

range

f (x) = \frac{1}{x ^{2} - 2}

f (x) \leq - \frac{1}{2} or f (x) > 0

Interval Notation

(- \infty, - \frac{1}{2}] \cup (0, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - 2

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

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

Discriminant

1 = y (x^{2} - 2) : 8 y^{2} + 4 y

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

Check if the range interval endpoints are included

y = - \frac{1}{2} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{1}{2} or f (x) > 0

y = 1 - x

f (x) = - (x + 2)^{3} + 1

f (x) = x^{2} + 6 x + 2

f (x) = lo g_{9} (- x)

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

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