What is the range of 1/(x^2-16) ?

The range of 1/(x^2-16) is f(x) 0

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range of 1/(x^2-16)

range

\frac{1}{x ^{2} - 16}

f (x) \leq - \frac{1}{16} or f (x) > 0

Interval Notation

(- \infty, - \frac{1}{16}] \cup (0, \infty)

Solution steps

Rewrite as

\frac{1}{x ^{2} - 16} = y

Multiply both sides by

x^{2} - 16

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

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

Discriminant

1 = y (x^{2} - 16) : 64 y^{2} + 4 y

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

Check if the range interval endpoints are included

y = - \frac{1}{16} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{1}{16} or f (x) > 0

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

g (x) = \frac{- x + 2}{7}

6 lo g_{5} (2 x - 6)

y = (x - 5)^{2} - 9

f (x) = - 2 x^{3} + 12 x^{2} + 1

What is the range of 1/(x^2-16) ?
The range of 1/(x^2-16) is f(x)<=-1/16 \lor f(x)>0