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

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

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range of 2/(x^2-2x-3)

range

\frac{2}{x ^{2} - 2 x - 3}

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

Interval Notation

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

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - 2 x - 3

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

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

Discriminant

2 = y (x^{2} - 2 x - 3) : 16 y^{2} + 8 y

16 y^{2} + 8 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

3 (x - 1)^{2} - 2

\frac{x - x ^{2}}{( x + 1 ) ^{2}}

y = x + 2 + 3

(2, 0), (0, - 2)

4 sin (3 π - 2 π x) - 7 π

What is the range of 2/(x^2-2x-3) ?
The range of 2/(x^2-2x-3) is f(x)<=-1/2 \lor f(x)>0