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

The range of (3x^2+6)/(x^2-2x-3) is f(x) = (3sqrt(33)+3)/8

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range of (3x^2+6)/(x^2-2x-3)

range

\frac{3 x ^{2} + 6}{x ^{2} - 2 x - 3}

f (x) \leq \frac{- 3 33 + 3}{8} or f (x) \geq \frac{3 33 + 3}{8}

Interval Notation

(- \infty, \frac{- 3 33 + 3}{8}] \cup [\frac{3 33 + 3}{8}, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} - 2 x - 3

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

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

Discriminant

3 x^{2} + 6 = y (x^{2} - 2 x - 3) : 16 y^{2} - 12 y - 72

16 y^{2} - 12 y - 72 \geq 0 : y \leq \frac{- 3 33 + 3}{8} or y \geq \frac{3 33 + 3}{8}

Check if the range interval endpoints are included

y = \frac{- 3 33 + 3}{8} :

Included

y = \frac{3 33 + 3}{8} :

Included

Therefore the range is

f (x) \leq \frac{- 3 33 + 3}{8} or f (x) \geq \frac{3 33 + 3}{8}

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

(0.3) (3)

f (x) = \frac{7 x}{9 x - 8}

f (x) = 2 x^{3} - 3 x^{2} - 36 x

(1.3) (5.1)

What is the range of (3x^2+6)/(x^2-2x-3) ?
The range of (3x^2+6)/(x^2-2x-3) is f(x)<= (-3sqrt(33)+3)/8 \lor f(x)>= (3sqrt(33)+3)/8