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

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

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range of 1/((x-2)^2)

range

\frac{1}{( x - 2 ) ^{2}}

f (x) > 0

Interval Notation

(0, \infty)

Solution steps

Rewrite as

\frac{1}{( x - 2 ) ^{2}} = y

Multiply both sides by

(x - 2)^{2}

1 = y (x - 2)^{2}

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

Discriminant

1 = y (x - 2)^{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

\frac{( 10 x ^{3} - 2 )}{2 x ^{3} + 5 x ^{2} + 8 x}

\frac{x ^{2} - 2 x + 1}{2 x ^{2} - x - 1}

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

f (x) = 2 x + 5

(- 4, 2), (2, 5)