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

The range of (9-x^2)/(2x^2) is f(x)>-1/2

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

range

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

f (x) > - \frac{1}{2}

Interval Notation

(- \frac{1}{2}, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x^{2}

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

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

Discriminant

\frac{9 - x ^{2}}{2} = y x^{2} : 72 y + 36

72 y + 36 \geq 0 : y \geq - \frac{1}{2}

Check if the range interval endpoints are included

y = - \frac{1}{2} :

Excluded

Therefore the range is

f (x) > - \frac{1}{2}

\frac{1}{x ^{2} - 4 x - 5}

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

f (x) = 3 x^{2} + \frac{4}{x ^{2}}

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

\frac{x ^{4}}{2} - \frac{3 x ^{2}}{2}