What is the range of (4x-8)/((x-4)(x+1)) ?

The range of (4x-8)/((x-4)(x+1)) is -infinity <f(x)<infinity

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range of (4x-8)/((x-4)(x+1))

range

\frac{4 x - 8}{( x - 4 ) ( x + 1 )}

- \infty < f (x) < \infty

Interval Notation

(- \infty, \infty)

Solution steps

Rewrite as

\frac{4 x - 8}{( x - 4 ) ( x + 1 )} = y

Multiply both sides by

(x - 4) (x + 1)

4 x - 8 = y (x - 4) (x + 1)

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

Discriminant

4 x - 8 = y (x - 4) (x + 1) : 25 y^{2} - 8 y + 16

25 y^{2} - 8 y + 16 \geq 0 :

True for all

y \in R

Therefore the range is

- \infty < f (x) < \infty

\frac{x ^{2} - 36}{x - 6}

7 x - y = - 7

f (x) = 6 x^{2} + 1

f (x) = - 3 (x + 2)^{2} + 5

\frac{x + 4}{x - 6}

What is the range of (4x-8)/((x-4)(x+1)) ?
The range of (4x-8)/((x-4)(x+1)) is -infinity <f(x)<infinity