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

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

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

range

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

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

Interval Notation

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

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - x

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

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

Discriminant

1 = y (x^{2} - 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 :

Excluded

Therefore the range is

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

y = - 2 x + 1

(- 6.6) - \frac{1}{4}

(2, - 1), (4, - 5)

\frac{3 + x}{x - 2}

f (x) = \frac{- 1}{x ^{2} + 5}

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