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

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

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range of f(x)= 2/(x^2-16)

range

f (x) = \frac{2}{x ^{2} - 16}

f (x) \leq - \frac{1}{8} or f (x) > 0

Interval Notation

(- \infty, - \frac{1}{8}] \cup (0, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - 16

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

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

Discriminant

2 = y (x^{2} - 16) : 64 y^{2} + 8 y

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

Check if the range interval endpoints are included

y = - \frac{1}{8} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{1}{8} or f (x) > 0

- 2 (3)^{x}

y = - \frac{2}{3} x - 5 (6.1)

sin (2 x)

f (x) = \frac{x + 6}{x - 9}

18500 (0.09 - x^{2})

What is the range of f(x)= 2/(x^2-16) ?
The range of f(x)= 2/(x^2-16) is f(x)<=-1/8 \lor f(x)>0