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

The range of y=(x^2)/(x^2-4) is f(x) 1

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

range

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

f (x) \leq 0 or f (x) > 1

Interval Notation

(- \infty, 0] \cup (1, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - 4

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

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

Discriminant

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

16 y^{2} - 16 y \geq 0 : y \leq 0 or y \geq 1

Check if the range interval endpoints are included

y = 0 :

Included

y = 1 :

Excluded

Therefore the range is

f (x) \leq 0 or f (x) > 1

4 x + 3 x - 12 = 0

5 x + y = 3

f (x) = T (x) = (x - 4) 3 + 6

f (x) = x^{4} - 2 x^{2} + 6

f (x) = ln (5 x - x^{2})

What is the range of y=(x^2)/(x^2-4) ?
The range of y=(x^2)/(x^2-4) is f(x)<= 0\lor f(x)>1