What is the range of f(x)=(2x^2)/(x^2-9) ?

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

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

range

f (x) = \frac{2 x ^{2}}{x ^{2} - 9}

f (x) \leq 0 or f (x) > 2

Interval Notation

(- \infty, 0] \cup (2, \infty)

Solution steps

Rewrite as

\frac{2 x ^{2}}{x ^{2} - 9} = y

Multiply both sides by

x^{2} - 9

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

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

Discriminant

2 x^{2} = y (x^{2} - 9) : 36 y^{2} - 72 y

36 y^{2} - 72 y \geq 0 : y \leq 0 or y \geq 2

Check if the range interval endpoints are included

y = 0 :

Included

y = 2 :

Excluded

Therefore the range is

f (x) \leq 0 or f (x) > 2

f (x) = (x - 7)^{2}

y = - x^{2} + 7 x + 1

2 cos (x) - sin (2 x)

f (x) = \frac{4}{x + 6}

f (x) = \frac{6 x - 6}{x + 2}

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