What is the range of f(x)= 1/((x-3)^2) ?

The range of f(x)= 1/((x-3)^2) is f(x)>0

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range of f(x)= 1/((x-3)^2)

range

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

f (x) > 0

Interval Notation

(0, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

(x - 3)^{2}

1 = y (x - 3)^{2}

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

Discriminant

1 = y (x - 3)^{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) = \frac{( x - 4 ) ( x - 3 )}{x + 3}

f (x) = - x^{2} + 14 x + 49

\frac{u + 1}{1 + \frac{1}{u + 1}}

y = \frac{3}{2} x, (2, 3)

- x^{2} - 36 + 6

What is the range of f(x)= 1/((x-3)^2) ?
The range of f(x)= 1/((x-3)^2) is f(x)>0