What is the range of f(x)= 1/(x^2-4) ?

The range of f(x)= 1/(x^2-4) is f(x) 0

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range of f(x)= 1/(x^2-4)

range

f (x) = \frac{1}{x ^{2} - 4}

f (x) \leq - \frac{1}{4} or f (x) > 0

Interval Notation

(- \infty, - \frac{1}{4}] \cup (0, \infty)

Solution steps

Rewrite as

\frac{1}{x ^{2} - 4} = y

Multiply both sides by

x^{2} - 4

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

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

Discriminant

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

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

Check if the range interval endpoints are included

y = - \frac{1}{4} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{1}{4} or f (x) > 0

f (x) = \frac{7}{x - 2}

\frac{2 x ^{2} - 6 x + 2}{( 2 x - 3 ) ^{2}}

(- 2, 4), (4, - 6)

y = \frac{2}{3} x + 1, (3, - 1)

cos (x) - 3

What is the range of f(x)= 1/(x^2-4) ?
The range of f(x)= 1/(x^2-4) is f(x)<=-1/4 \lor f(x)>0