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

The range of f(x)=(x^2)/(x-9) is f(x) = 36

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

range

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

f (x) \leq 0 or f (x) \geq 36

Interval Notation

(- \infty, 0] \cup [36, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x - 9

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

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

Discriminant

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

y^{2} - 36 y \geq 0 : y \leq 0 or y \geq 36

Check if the range interval endpoints are included

y = 0 :

Included

y = 36 :

Included

Therefore the range is

f (x) \leq 0 or f (x) \geq 36

(- 2.6) - 3

1 + 3.22 lo g_{34} (x)

f (x) = - x + 4

\frac{1}{x + 1}

What is the range of f(x)=(x^2)/(x-9) ?
The range of f(x)=(x^2)/(x-9) is f(x)<= 0\lor f(x)>= 36