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

The range of y= 1/(x^2) is f(x)>0

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range of y= 1/(x^2)

range

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

f (x) > 0

Interval Notation

(0, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x^{2}

1 = y x^{2}

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

Discriminant

1 = y x^{2} : 4 y

4 y \geq 0 : y \geq 0

Check if the range interval endpoints are included

y = 0 :

Excluded

Therefore the range is

f (x) > 0

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

\frac{sec ^{2} ( x )}{3 ln ( 2 ) tan ( x )}

e^{- x} - 1

sin (12 x)

(- 9, 1), (- 1, 9)