What is the range of y=((x^2))/(x^2-16) ?

The range of y=((x^2))/(x^2-16) is f(x) 1

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range of y=((x^2))/(x^2-16)

range

y = \frac{( x ^{2} )}{x ^{2} - 16}

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

Interval Notation

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

Solution steps

Rewrite as

\frac{x ^{2}}{x ^{2} - 16} = y

Multiply both sides by

x^{2} - 16

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

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

Discriminant

x^{2} = y (x^{2} - 16) : 64 y^{2} - 64 y

64 y^{2} - 64 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

f (x) = \frac{2 x + 3}{5 x + 4}

\frac{7}{x + 3}

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

x^{3} - 12 x + 3

4 x^{7} + 6

What is the range of y=((x^2))/(x^2-16) ?
The range of y=((x^2))/(x^2-16) is f(x)<= 0\lor f(x)>1