What is the range of f(x)= 4/(x^2-4x) ?

The range of f(x)= 4/(x^2-4x) is f(x) 0

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range of f(x)= 4/(x^2-4x)

range

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

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

Interval Notation

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

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - 4 x

4 = y (x^{2} - 4 x)

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

Discriminant

4 = y (x^{2} - 4 x) : 16 y^{2} + 16 y

16 y^{2} + 16 y \geq 0 : y \leq - 1 or y \geq 0

Check if the range interval endpoints are included

y = - 1 :

Included

y = 0 :

Excluded

Therefore the range is

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

f (x) = \frac{5 x + 1}{9 x - 2}

f (x) = \frac{7 x}{x - 2}

\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2}

f (x) = \frac{1}{2} x - 8

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