What is the range of f(x)=(x^2)/(1-x) ?

The range of f(x)=(x^2)/(1-x) is f(x) = 0

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

range

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

f (x) \leq - 4 or f (x) \geq 0

Interval Notation

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

Solution steps

Rewrite as

\frac{x ^{2}}{1 - x} = y

Multiply both sides by

1 - x

x^{2} = y (1 - x)

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

Discriminant

x^{2} = y (1 - x) : y^{2} + 4 y

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

Check if the range interval endpoints are included

y = - 4 :

Included

y = 0 :

Included

Therefore the range is

f (x) \leq - 4 or f (x) \geq 0

f (x) = \frac{x ^{2} + 2 x - 15}{x ^{2} - 4}

x^{2} - x - 1

(- 2.1) - \frac{1}{2}

f (x) = \frac{2 x ^{2} + 10 x}{3 x + 15}

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

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