What is the domain of f(x)=sqrt(2x^2-x-6+2) ?

The domain of f(x)=sqrt(2x^2-x-6+2) is x = (sqrt(33)+1)/4

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(2x^2-x-6+2)

domain

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

x \leq \frac{- 33 + 1}{4} or x \geq \frac{33 + 1}{4}

Interval Notation

(- \infty, \frac{- 33 + 1}{4}] \cup [\frac{33 + 1}{4}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq \frac{- 33 + 1}{4} or x \geq \frac{33 + 1}{4}

The function domain

x \leq \frac{- 33 + 1}{4} or x \geq \frac{33 + 1}{4}

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

f (x) = \frac{x + 2}{x + 1}

y = 3 - x

f (x) = lo g_{10} (5 x + 35)

f (x) = 5 - t

What is the domain of f(x)=sqrt(2x^2-x-6+2) ?
The domain of f(x)=sqrt(2x^2-x-6+2) is x<= (-sqrt(33)+1)/4 \lor x>= (sqrt(33)+1)/4