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

The range of (3x^2-3)/(x^2+x-6) is f(x) = (12sqrt(6)+42)/(25)

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

range

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

f (x) \leq \frac{- 12 6 + 42}{25} or f (x) \geq \frac{12 6 + 42}{25}

Interval Notation

(- \infty, \frac{- 12 6 + 42}{25}] \cup [\frac{12 6 + 42}{25}, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

x^{2} + x - 6

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

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

Discriminant

3 x^{2} - 3 = y (x^{2} + x - 6) : 25 y^{2} - 84 y + 36

25 y^{2} - 84 y + 36 \geq 0 : y \leq \frac{- 12 6 + 42}{25} or y \geq \frac{12 6 + 42}{25}

Check if the range interval endpoints are included

y = \frac{- 12 6 + 42}{25} :

Included

y = \frac{12 6 + 42}{25} :

Included

Therefore the range is

f (x) \leq \frac{- 12 6 + 42}{25} or f (x) \geq \frac{12 6 + 42}{25}

\frac{12 x + 35}{x ( x + 7 )}

f (x) = x^{3} - 15 x^{2} + 75 x - 1

\frac{1}{4} x^{2} - 2 x + 15

\frac{x + 4}{x ^{2} + 25}

12 x^{5} + 15 x^{4} - 240 x^{3} + 1

What is the range of (3x^2-3)/(x^2+x-6) ?
The range of (3x^2-3)/(x^2+x-6) is f(x)<= (-12sqrt(6)+42)/(25)\lor f(x)>= (12sqrt(6)+42)/(25)