What is the domain of (1+5x)^{1/x+2} ?

The domain of (1+5x)^{1/x+2} is -1/5 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

domain of (1+5x)^{1/x+2}

domain

(1 + 5 x)^{\frac{1}{x} + 2}

- \frac{1}{5} < x < 0 or x > 0

Interval Notation

(- \frac{1}{5}, 0) \cup (0, \infty)

Solution steps

Check exponents domain:

x > - \frac{1}{5}

Find undefined (singularity) points:

x = 0

Combine real regions and undefined points for final function domain

- \frac{1}{5} < x < 0 or x > 0

f (x) = \frac{- 4 x ^{2}}{( x - 4 ) ( x + 2 )}

f (x) = \frac{x - 5}{3 x - 18}

f (x) = x^{2} + 300 x - 100000

y = x - 8

f (x) = - 2 1 - x

What is the domain of (1+5x)^{1/x+2} ?
The domain of (1+5x)^{1/x+2} is -1/5 <x<0\lor x>0