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

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

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

range

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

f (x) \leq - \frac{3}{4} or f (x) > 0

Interval Notation

(- \infty, - \frac{3}{4}] \cup (0, \infty)

Solution steps

Rewrite as

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

Multiply both sides by

(x - 2) (x + 2)

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

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

Discriminant

3 = y (x - 2) (x + 2) : 16 y^{2} + 12 y

16 y^{2} + 12 y \geq 0 : y \leq - \frac{3}{4} or y \geq 0

Check if the range interval endpoints are included

y = - \frac{3}{4} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{3}{4} or f (x) > 0

sin^{2} (x)

f (x) = x^{3} - 12 x + 5

f (x) = x^{2} + 5 x, x > 0

2 x - y = 6

m = 1, (- 4, 7)

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