What is the range of (x^2+5)/(x-1) ?

The range of (x^2+5)/(x-1) is f(x) = 2sqrt(6)+2

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range of (x^2+5)/(x-1)

range

\frac{x ^{2} + 5}{x - 1}

f (x) \leq - 26 + 2 or f (x) \geq 26 + 2

Interval Notation

(- \infty, - 26 + 2] \cup [26 + 2, \infty)

Solution steps

Rewrite as

\frac{x ^{2} + 5}{x - 1} = y

Multiply both sides by

x - 1

x^{2} + 5 = y (x - 1)

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

Discriminant

x^{2} + 5 = y (x - 1) : y^{2} - 20 - 4 y

y^{2} - 20 - 4 y \geq 0 : y \leq - 26 + 2 or y \geq 26 + 2

Check if the range interval endpoints are included

y = - 26 + 2 :

Included

y = 26 + 2 :

Included

Therefore the range is

f (x) \leq - 26 + 2 or f (x) \geq 26 + 2

- 4 t^{2} - 8 t + 6.8

ln (\frac{2 x + 3}{6 - x})

f (x) = - 5 x - 5

f (x) = \frac{6}{x - 9}

y = 2 x^{2} + 5 x - 7

What is the range of (x^2+5)/(x-1) ?
The range of (x^2+5)/(x-1) is f(x)<=-2sqrt(6)+2\lor f(x)>= 2sqrt(6)+2