What is the range of 4/(t^2-9) ?

The range of 4/(t^2-9) is f(t) 0

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range of 4/(t^2-9)

range

\frac{4}{t ^{2} - 9}

f (t) \leq - \frac{4}{9} or f (t) > 0

Interval Notation

(- \infty, - \frac{4}{9}] \cup (0, \infty)

Solution steps

Rewrite as

\frac{4}{t ^{2} - 9} = y

Multiply both sides by

t^{2} - 9

4 = y (t^{2} - 9)

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

Discriminant

4 = y (t^{2} - 9) : 36 y^{2} + 16 y

36 y^{2} + 16 y \geq 0 : y \leq - \frac{4}{9} or y \geq 0

Check if the range interval endpoints are included

y = - \frac{4}{9} :

Included

y = 0 :

Excluded

Therefore the range is

f (t) \leq - \frac{4}{9} or f (t) > 0

x + 3 y = - 6

(8, 10), (2, 6)

(x + 2) (x - 3)

f (x) = \frac{x + 5}{x ^{2} + 3}

f (x) = 1

What is the range of 4/(t^2-9) ?
The range of 4/(t^2-9) is f(t)<=-4/9 \lor f(t)>0