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

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

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range of f(x)=(x^2+5)/(x+1)

range

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

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

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

- 8 x - 3 y = 9

(1.172013) (3.252015)

x^{3} - 9 x^{2} + 15 x + 8

f (x) = \frac{x ^{2}}{x - 8}

3 (x - 9)

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