What is the range of f(x)= 9/((x-2)^2) ?

The range of f(x)= 9/((x-2)^2) is f(x)>0

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range of f(x)= 9/((x-2)^2)

range

f (x) = \frac{9}{( x - 2 ) ^{2}}

f (x) > 0

Interval Notation

(0, \infty)

Solution steps

Rewrite as

\frac{9}{( x - 2 ) ^{2}} = y

Multiply both sides by

(x - 2)^{2}

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

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

Discriminant

9 = y (x - 2)^{2} : 36 y

36 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) = \frac{1 - 3 x}{2 + x}

- 2 + sin (x + \frac{π}{3})

f (x) = \frac{x + 4}{x ^{3} - 12 x ^{2} + 32 x}

f (x) = 2 x^{2} - x + 7

(- 10, 21), (8, - 6)

What is the range of f(x)= 9/((x-2)^2) ?
The range of f(x)= 9/((x-2)^2) is f(x)>0