What is the range of f(x)= 2/(t^2-9) ?

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

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range of f(x)= 2/(t^2-9)

range

f (x) = \frac{2}{t ^{2} - 9}

f (x) \leq - \frac{2}{9} or f (x) > 0

Interval Notation

(- \infty, - \frac{2}{9}] \cup (0, \infty)

Solution steps

Rewrite as

\frac{2}{t ^{2} - 9} = x

Multiply both sides by

t^{2} - 9

2 = x (t^{2} - 9)

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

Discriminant

2 = x (t^{2} - 9) : 36 x^{2} + 8 x

36 x^{2} + 8 x \geq 0 : x \leq - \frac{2}{9} or x \geq 0

Check if the range interval endpoints are included

x = - \frac{2}{9} :

Included

x = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{2}{9} or f (x) > 0

f (x) = 2^{x^{5} - 1}

- 4 x^{5} - 4

(- 18, - 20), (- 13, - 15)

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

3 x + 1

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