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

The range of f(x)=(x^2-x-2)/(x-3) is f(x) = 9

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

range

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

f (x) \leq 1 or f (x) \geq 9

Interval Notation

(- \infty, 1] \cup [9, \infty)

Solution steps

Rewrite as

\frac{x ^{2} - x - 2}{x - 3} = y

Multiply both sides by

x - 3

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

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

Discriminant

x^{2} - x - 2 = y (x - 3) : y^{2} - 10 y + 9

y^{2} - 10 y + 9 \geq 0 : y \leq 1 or y \geq 9

Check if the range interval endpoints are included

y = 1 :

Included

y = 9 :

Included

Therefore the range is

f (x) \leq 1 or f (x) \geq 9

f (x) = 6 x^{3} + 8 x^{2}

f (x) = - 5 x^{2} - 30 x - 42

(- 1, 4), (1, - 6)

f (x) = \frac{x ^{3} + 5}{x ^{5} + 2}

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

What is the range of f(x)=(x^2-x-2)/(x-3) ?
The range of f(x)=(x^2-x-2)/(x-3) is f(x)<= 1\lor f(x)>= 9