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

The range of y=(x^2+x+2)/(x-1) is f(x) = 7

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range of y=(x^2+x+2)/(x-1)

range

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

f (x) \leq - 1 or f (x) \geq 7

Interval Notation

(- \infty, - 1] \cup [7, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x - 1

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

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

Discriminant

x^{2} + x + 2 = y (x - 1) : y^{2} - 6 y - 7

y^{2} - 6 y - 7 \geq 0 : y \leq - 1 or y \geq 7

Check if the range interval endpoints are included

y = - 1 :

Included

y = 7 :

Included

Therefore the range is

f (x) \leq - 1 or f (x) \geq 7

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

f (x) = 9.5 x^{2} - x + 2.6

\frac{x ^{3}}{3} + 2 x^{2} + 4 x - 1

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

f (x) = x^{3} - 3 x + 2

What is the range of y=(x^2+x+2)/(x-1) ?
The range of y=(x^2+x+2)/(x-1) is f(x)<=-1\lor f(x)>= 7