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

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

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

range

f (x) = \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

(0, 7), (4, 6)

f (x) = - x + 5

x + 5 y = - 5

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

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