What is the range of (7x-21)/((x-7)(x+1)) ?

The range of (7x-21)/((x-7)(x+1)) is -infinity <f(x)<infinity

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range of (7x-21)/((x-7)(x+1))

range

\frac{7 x - 21}{( x - 7 ) ( x + 1 )}

- \infty < f (x) < \infty

Interval Notation

(- \infty, \infty)

Solution steps

Rewrite as

\frac{7 x - 21}{( x - 7 ) ( x + 1 )} = y

Multiply both sides by

(x - 7) (x + 1)

7 x - 21 = y (x - 7) (x + 1)

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

Discriminant

7 x - 21 = y (x - 7) (x + 1) : 64 y^{2} + 49

64 y^{2} + 49 \geq 0 :

True for all

y \in R

Therefore the range is

- \infty < f (x) < \infty

f (x) = \frac{( x + 5 )}{x + 10}

f (x) = \frac{3 - 2 x}{x ^{2} - 64}

x^{2} + 6 x + 9

f (x) = (x - 1) (2 x + 3)

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

What is the range of (7x-21)/((x-7)(x+1)) ?
The range of (7x-21)/((x-7)(x+1)) is -infinity <f(x)<infinity