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

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

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range of 1/(x^2-x)

range

\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

f (x) = \frac{2 x + 1}{7}

f (x) = x e^{2 x}

f (x) = 8 (x - 3) + 4

f (x) = - 2 x + 25

\frac{x}{1 - ln ( x - 9 )}