What is the range of y=(x-1)/(x^2-9) ?

The range of y=(x-1)/(x^2-9) is -infinity <f(x)<infinity

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range of y=(x-1)/(x^2-9)

range

y = \frac{x - 1}{x ^{2} - 9}

- \infty < f (x) < \infty

Solution steps

Rewrite as

\frac{x - 1}{x ^{2} - 9} = y

Multiply both sides by

x^{2} - 9

x - 1 = y (x^{2} - 9)

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

Discriminant

x - 1 = y (x^{2} - 9) : 1 + 36 y^{2} - 4 y

1 + 36 y^{2} - 4 y \geq 0 :

True for all

y \in R

Therefore the range is

- \infty < f (x) < \infty

y = \frac{5}{x}

\frac{e ^{x}}{x ^{2}}

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

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

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

What is the range of y=(x-1)/(x^2-9) ?
The range of y=(x-1)/(x^2-9) is -infinity <f(x)<infinity