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

The range of ((x^2+5))/(2x^2-x-1) is f(x) = (2sqrt(14))/3-2

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

range

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

f (x) \leq - \frac{2 14}{3} - 2 or f (x) \geq \frac{2 14}{3} - 2

Interval Notation

(- \infty, - \frac{2 14}{3} - 2] \cup [\frac{2 14}{3} - 2, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

2 x^{2} - x - 1

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

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

Discriminant

x^{2} + 5 = y (2 x^{2} - x - 1) : 9 y^{2} + 36 y - 20

9 y^{2} + 36 y - 20 \geq 0 : y \leq - \frac{2 14}{3} - 2 or y \geq \frac{2 14}{3} - 2

Check if the range interval endpoints are included

y = - \frac{2 14}{3} - 2 :

Included

y = \frac{2 14}{3} - 2 :

Included

Therefore the range is

f (x) \leq - \frac{2 14}{3} - 2 or f (x) \geq \frac{2 14}{3} - 2

x^{2} + 2 x - 1

f (x) = 12 + 4 x - x^{2}, 0 \leq x \leq 5

f (x) = 6

6 x - 10 y = - 3

f (x) = 7 - 2 x

What is the range of ((x^2+5))/(2x^2-x-1) ?
The range of ((x^2+5))/(2x^2-x-1) is f(x)<=-(2sqrt(14))/3-2\lor f(x)>= (2sqrt(14))/3-2