What is the domain of f(x)=sqrt(8x-1)+5x^2 ?

The domain of f(x)=sqrt(8x-1)+5x^2 is x>= 1/8

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(8x-1)+5x^2

domain

f (x) = 8 x - 1 + 5 x^{2}

x \geq \frac{1}{8}

Interval Notation

[\frac{1}{8}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq \frac{1}{8}

The function domain

x \geq \frac{1}{8}

- \frac{1}{2} x

1 + \frac{1}{x}

\frac{5 x - 3}{3 x + 2}

sin (x) + cos (x)

f (x) = \frac{1}{2} ∣ x ∣

What is the domain of f(x)=sqrt(8x-1)+5x^2 ?
The domain of f(x)=sqrt(8x-1)+5x^2 is x>= 1/8