What is the range of (x-8)/(2x^2) ?

The range of (x-8)/(2x^2) is f(x)<= 1/64

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range of (x-8)/(2x^2)

range

\frac{x - 8}{2 x ^{2}}

f (x) \leq \frac{1}{64}

Interval Notation

(- \infty, \frac{1}{64}]

Solution steps

Rewrite as

\frac{x - 8}{2 x ^{2}} = y

Multiply both sides by

x^{2}

\frac{x - 8}{2} = y x^{2}

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

Discriminant

\frac{x - 8}{2} = y x^{2} : 1 - 64 y

1 - 64 y \geq 0 : y \leq \frac{1}{64}

Check if the range interval endpoints are included

y = \frac{1}{64} :

Included

Therefore the range is

f (x) \leq \frac{1}{64}

f (x) = x - 1

- 2 x + 2

5^{x}

f (x) = \frac{x + 10}{x - 5}

f (x) = arccsc (x)