What is the range of 1/(2x^2-x-6) ?

The range of 1/(2x^2-x-6) is f(x) 0

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range of 1/(2x^2-x-6)

range

\frac{1}{2 x ^{2} - x - 6}

f (x) \leq - \frac{8}{49} or f (x) > 0

Interval Notation

(- \infty, - \frac{8}{49}] \cup (0, \infty)

Solution steps

Rewrite as

\frac{1}{2 x ^{2} - x - 6} = y

Multiply both sides by

2 x^{2} - x - 6

1 = y (2 x^{2} - x - 6)

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

Discriminant

1 = y (2 x^{2} - x - 6) : 49 y^{2} + 8 y

49 y^{2} + 8 y \geq 0 : y \leq - \frac{8}{49} or y \geq 0

Check if the range interval endpoints are included

y = - \frac{8}{49} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{8}{49} or f (x) > 0

y = tan (x)

x^{3} - 12 x^{2} - 27 x + 2

\frac{2 + x}{x - 5}

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

f (x) = 2 x^{2} + 4 x - 3

What is the range of 1/(2x^2-x-6) ?
The range of 1/(2x^2-x-6) is f(x)<=-8/49 \lor f(x)>0