What is the range of (x^2+x+1)/x ?

The range of (x^2+x+1)/x is f(x) = 3

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range of (x^2+x+1)/x

range

\frac{x ^{2} + x + 1}{x}

f (x) \leq - 1 or f (x) \geq 3

Interval Notation

(- \infty, - 1] \cup [3, \infty)

Solution steps

Rewrite as

\frac{x ^{2} + x + 1}{x} = y

Multiply both sides by

x

x^{2} + x + 1 = y x

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

Discriminant

x^{2} + x + 1 = y x : y^{2} - 2 y - 3

y^{2} - 2 y - 3 \geq 0 : y \leq - 1 or y \geq 3

Check if the range interval endpoints are included

y = - 1 :

Included

y = 3 :

Included

Therefore the range is

f (x) \leq - 1 or f (x) \geq 3

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

x^{3} + 2 x^{2} + x - 2

f (x) = (x^{2} - 4)^{2}

\frac{x ^{2} - 9}{x ^{2} - 4 x + 3}

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

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