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

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

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

range

f (x) = \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

x = - 5, (6, - 6)

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

1 + \frac{1}{x - 1}

3 + (8 + x)^{\frac{1}{2}}

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