What is the range of (x^2-5)/(7x^2) ?

The range of (x^2-5)/(7x^2) is f(x)< 1/7

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range of (x^2-5)/(7x^2)

range

\frac{x ^{2} - 5}{7 x ^{2}}

f (x) < \frac{1}{7}

Interval Notation

(- \infty, \frac{1}{7})

Solution steps

Rewrite as

\frac{x ^{2} - 5}{7 x ^{2}} = y

Multiply both sides by

x^{2}

\frac{x ^{2} - 5}{7} = y x^{2}

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

Discriminant

\frac{x ^{2} - 5}{7} = y x^{2} : - 140 y + 20

- 140 y + 20 \geq 0 : y \leq \frac{1}{7}

Check if the range interval endpoints are included

y = \frac{1}{7} :

Excluded

Therefore the range is

f (x) < \frac{1}{7}

f (x) = \frac{x ^{2} - 1}{x ^{2} + 2 x - 3}

f (x) = \frac{3}{x ^{2} + 6 x - 4}

f (x) = \frac{2}{1 - e ^{x + 2}}

f (x) = 9 - x^{2} \cdot x + 1