What is the domain of f(x)=(\sqrt[4]{2x+9})^{-3} ?

The domain of f(x)=(\sqrt[4]{2x+9})^{-3} is x>-9/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

domain of f(x)=(\sqrt[4]{2x+9})^{-3}

domain

f (x) = (4 2 x + 9)^{- 3}

x > - \frac{9}{2}

Interval Notation

(- \frac{9}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - \frac{9}{2}

Find undefined (singularity) points:

x = - \frac{9}{2}

Combine real regions and undefined points for final function domain

x > - \frac{9}{2}

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

f (x) = 2 x^{2} - 6 x - 8

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

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

f (x) = \frac{x}{x - 9}

What is the domain of f(x)=(\sqrt[4]{2x+9})^{-3} ?
The domain of f(x)=(\sqrt[4]{2x+9})^{-3} is x>-9/2