What is the range of f(x)= 1/((x-8)^2) ?

The range of f(x)= 1/((x-8)^2) is f(x)>0

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range of f(x)= 1/((x-8)^2)

range

f (x) = \frac{1}{( x - 8 ) ^{2}}

f (x) > 0

Interval Notation

(0, \infty)

Solution steps

Rewrite as

\frac{1}{( x - 8 ) ^{2}} = y

Multiply both sides by

(x - 8)^{2}

1 = y (x - 8)^{2}

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

Discriminant

1 = y (x - 8)^{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) = x^{3} + 10

f (x) = 3 lo g_{2} (x)

e^{x - 1}

f (x) = \frac{1}{5} x + 2

f (x) = 15 x - 15

What is the range of f(x)= 1/((x-8)^2) ?
The range of f(x)= 1/((x-8)^2) is f(x)>0