What is the range of f(t)= 2/(t^2-16) ?

The range of f(t)= 2/(t^2-16) is f(t) 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

range of f(t)= 2/(t^2-16)

range

f (t) = \frac{2}{t ^{2} - 16}

f (t) \leq - \frac{1}{8} or f (t) > 0

Interval Notation

(- \infty, - \frac{1}{8}] \cup (0, \infty)

Solution steps

Rewrite as

\frac{2}{t ^{2} - 16} = y

Multiply both sides by

t^{2} - 16

2 = y (t^{2} - 16)

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

Discriminant

2 = y (t^{2} - 16) : 64 y^{2} + 8 y

64 y^{2} + 8 y \geq 0 : y \leq - \frac{1}{8} or y \geq 0

Check if the range interval endpoints are included

y = - \frac{1}{8} :

Included

y = 0 :

Excluded

Therefore the range is

f (t) \leq - \frac{1}{8} or f (t) > 0

8 (q - 13)

f (x) = 3 - 2 x - x^{2}

f (x) = - x^{3} + 7 x^{2} - 12 x

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

f (x) = \frac{1}{( x - 3 ) ( x - 5 )}

What is the range of f(t)= 2/(t^2-16) ?
The range of f(t)= 2/(t^2-16) is f(t)<=-1/8 \lor f(t)>0