What is the range of (x^2+4x+6)/(3x^2+12x+12) ?

The range of (x^2+4x+6)/(3x^2+12x+12) is f(x)> 1/3

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range of (x^2+4x+6)/(3x^2+12x+12)

range

\frac{x ^{2} + 4 x + 6}{3 x ^{2} + 12 x + 12}

f (x) > \frac{1}{3}

Interval Notation

(\frac{1}{3}, \infty)

Solution steps

Rewrite as

\frac{x ^{2} + 4 x + 6}{3 x ^{2} + 12 x + 12} = y

Multiply both sides by

3 x^{2} + 12 x + 12

x^{2} + 4 x + 6 = y (3 x^{2} + 12 x + 12)

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

Discriminant

x^{2} + 4 x + 6 = y (3 x^{2} + 12 x + 12) : 24 y - 8

24 y - 8 \geq 0 : y \geq \frac{1}{3}

Check if the range interval endpoints are included

y = \frac{1}{3} :

Excluded

Therefore the range is

f (x) > \frac{1}{3}

y = - x - 4

x + x + 9

2 x - 2 y = 14

(- 5, 4), (1, 6)

f (x) = 8

What is the range of (x^2+4x+6)/(3x^2+12x+12) ?
The range of (x^2+4x+6)/(3x^2+12x+12) is f(x)> 1/3