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}

- \frac{9}{2 x ^{\frac{3}{2}}}

f (x) = - 10 lo g_{4} (x)

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

m = - 2, (4, 0)

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

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