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

The range of (2x^2)/(x^2-x-2) is f(x) = 16/9

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

range

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

f (x) \leq 0 or f (x) \geq \frac{16}{9}

Interval Notation

(- \infty, 0] \cup [\frac{16}{9}, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - x - 2

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

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

Discriminant

2 x^{2} = y (x^{2} - x - 2) : 9 y^{2} - 16 y

9 y^{2} - 16 y \geq 0 : y \leq 0 or y \geq \frac{16}{9}

Check if the range interval endpoints are included

y = 0 :

Included

y = \frac{16}{9} :

Included

Therefore the range is

f (x) \leq 0 or f (x) \geq \frac{16}{9}

What is the range of (2x^2)/(x^2-x-2) ?
The range of (2x^2)/(x^2-x-2) is f(x)<= 0\lor f(x)>= 16/9