What is the range of f(x)=(x^2+6)/(x+3) ?

The range of f(x)=(x^2+6)/(x+3) is f(x) = 2sqrt(15)-6

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range of f(x)=(x^2+6)/(x+3)

range

f (x) = \frac{x ^{2} + 6}{x + 3}

f (x) \leq - 215 - 6 or f (x) \geq 215 - 6

Interval Notation

(- \infty, - 215 - 6] \cup [215 - 6, \infty)

Solution steps

Rewrite as

\frac{x ^{2} + 6}{x + 3} = y

Multiply both sides by

x + 3

x^{2} + 6 = y (x + 3)

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

Discriminant

x^{2} + 6 = y (x + 3) : y^{2} - 24 + 12 y

y^{2} - 24 + 12 y \geq 0 : y \leq - 215 - 6 or y \geq 215 - 6

Check if the range interval endpoints are included

y = - 215 - 6 :

Included

y = 215 - 6 :

Included

Therefore the range is

f (x) \leq - 215 - 6 or f (x) \geq 215 - 6

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

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

f (x) = 3 x^{4}

- 3 sin (2 π x + 2)

y = - tan (x - \frac{π}{3})

What is the range of f(x)=(x^2+6)/(x+3) ?
The range of f(x)=(x^2+6)/(x+3) is f(x)<=-2sqrt(15)-6\lor f(x)>= 2sqrt(15)-6