What is the range of f(x)=(4x^2)/(x^2-9) ?

The range of f(x)=(4x^2)/(x^2-9) is f(x) 4

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range of f(x)=(4x^2)/(x^2-9)

range

f (x) = \frac{4 x ^{2}}{x ^{2} - 9}

f (x) \leq 0 or f (x) > 4

Interval Notation

(- \infty, 0] \cup (4, \infty)

Solution steps

Rewrite as

\frac{4 x ^{2}}{x ^{2} - 9} = y

Multiply both sides by

x^{2} - 9

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

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

Discriminant

4 x^{2} = y (x^{2} - 9) : 36 y^{2} - 144 y

36 y^{2} - 144 y \geq 0 : y \leq 0 or y \geq 4

Check if the range interval endpoints are included

y = 0 :

Included

y = 4 :

Excluded

Therefore the range is

f (x) \leq 0 or f (x) > 4

2 x + 5 y + 20 = 0

f (x) = \frac{x ^{2} + 5 x - 6}{x ^{2} - 5 x - 6}

\frac{- 2 - 5 x}{3 x - 1}

f (x) = - \frac{1}{x ^{2} + 7}

f (x) = \frac{t}{t + 1}

What is the range of f(x)=(4x^2)/(x^2-9) ?
The range of f(x)=(4x^2)/(x^2-9) is f(x)<= 0\lor f(x)>4