What is the range of f(x)= 3/(x^2-16) ?

The range of f(x)= 3/(x^2-16) is f(x) 0

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range of f(x)= 3/(x^2-16)

range

f (x) = \frac{3}{x ^{2} - 16}

f (x) \leq - \frac{3}{16} or f (x) > 0

Interval Notation

(- \infty, - \frac{3}{16}] \cup (0, \infty)

Solution steps

Rewrite as

\frac{3}{x ^{2} - 16} = y

Multiply both sides by

x^{2} - 16

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

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

Discriminant

3 = y (x^{2} - 16) : 64 y^{2} + 12 y

64 y^{2} + 12 y \geq 0 : y \leq - \frac{3}{16} or y \geq 0

Check if the range interval endpoints are included

y = - \frac{3}{16} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{3}{16} or f (x) > 0

x^{2} + 6 x - 1

f (x) = \frac{x + 6}{x ( x + 11 )}

g (t) = \frac{16}{1 + 3 ^{- t}}

\frac{3 x + 4}{6 x + 1}

f (x) = (x^{2} - 2 x - 8)^{\frac{1}{3}}

What is the range of f(x)= 3/(x^2-16) ?
The range of f(x)= 3/(x^2-16) is f(x)<=-3/16 \lor f(x)>0