What is the range of 1/((x+2)(x-3)) ?

The range of 1/((x+2)(x-3)) is f(x) 0

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range of 1/((x+2)(x-3))

range

\frac{1}{( x + 2 ) ( x - 3 )}

f (x) \leq - \frac{4}{25} or f (x) > 0

Interval Notation

(- \infty, - \frac{4}{25}] \cup (0, \infty)

Solution steps

Rewrite as

\frac{1}{( x + 2 ) ( x - 3 )} = y

Multiply both sides by

(x + 2) (x - 3)

1 = y (x + 2) (x - 3)

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

Discriminant

1 = y (x + 2) (x - 3) : 25 y^{2} + 4 y

25 y^{2} + 4 y \geq 0 : y \leq - \frac{4}{25} or y \geq 0

Check if the range interval endpoints are included

y = - \frac{4}{25} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{4}{25} or f (x) > 0

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

y = x^{2} - 6

(- 7, - 4), (- 1, 6)

(0, 1), (- 5, 4)

- 5 csc (π x)

What is the range of 1/((x+2)(x-3)) ?
The range of 1/((x+2)(x-3)) is f(x)<=-4/25 \lor f(x)>0