What is the range of y=(4x^2)/(x^2-2x-3) ?

The range of y=(4x^2)/(x^2-2x-3) is f(x) = 3

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range of y=(4x^2)/(x^2-2x-3)

range

y = \frac{4 x ^{2}}{x ^{2} - 2 x - 3}

f (x) \leq 0 or f (x) \geq 3

Interval Notation

(- \infty, 0] \cup [3, \infty)

Solution steps

Rewrite as

\frac{4 x ^{2}}{x ^{2} - 2 x - 3} = y

Multiply both sides by

x^{2} - 2 x - 3

4 x^{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

4 x^{2} = y (x^{2} - 2 x - 3) : 16 y^{2} - 48 y

16 y^{2} - 48 y \geq 0 : y \leq 0 or y \geq 3

Check if the range interval endpoints are included

y = 0 :

Included

y = 3 :

Included

Therefore the range is

f (x) \leq 0 or f (x) \geq 3

x^{3} - 12 x + 3

4 x^{7} + 6

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

\frac{1}{x ^{2} - 4 x - 5}

f (x) = - \frac{8}{x}

What is the range of y=(4x^2)/(x^2-2x-3) ?
The range of y=(4x^2)/(x^2-2x-3) is f(x)<= 0\lor f(x)>= 3