What is the range of (2x)/(x^2-9) ?

The range of (2x)/(x^2-9) is -infinity <f(x)<infinity

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range of (2x)/(x^2-9)

range

\frac{2 x}{x ^{2} - 9}

- \infty < f (x) < \infty

Interval Notation

(- \infty, \infty)

Solution steps

Rewrite as

\frac{2 x}{x ^{2} - 9} = y

Multiply both sides by

x^{2} - 9

2 x = y (x^{2} - 9)

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

Discriminant

2 x = y (x^{2} - 9) : 4 + 36 y^{2}

4 + 36 y^{2} \geq 0 :

True for all

y \in R

Therefore the range is

- \infty < f (x) < \infty

f (x) = cos (7 x)

x + 2 - 2

\frac{1}{x + 4} + 3

y = - 2 x + 5

\frac{490 \cdot e ^{3 x}}{4 + e ^{3 x}}

What is the range of (2x)/(x^2-9) ?
The range of (2x)/(x^2-9) is -infinity <f(x)<infinity