What is the domain of f(x)=sqrt(10-x)+sqrt(x^2-9) ?

The domain of f(x)=sqrt(10-x)+sqrt(x^2-9) is x<=-3\lor 3<= x<= 10

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(10-x)+sqrt(x^2-9)

domain

f (x) = 10 - x + x^{2} - 9

x \leq - 3 or 3 \leq x \leq 10

Interval Notation

(- \infty, - 3] \cup [3, 10]

Solution steps

Find non-negative values for radicals:

x \leq - 3 or 3 \leq x \leq 10

The function domain

x \leq - 3 or 3 \leq x \leq 10

\frac{1}{x ^{4} + 4 x ^{2} - 1}

\frac{- x + 1}{3 x + 6}

f (x) = \frac{2 x - 1}{2 x - 6}

\frac{4 - x}{x ^{2} - 16}

f (x) = \frac{x ^{2} - 4}{x ^{2} - 2 x + 1}

What is the domain of f(x)=sqrt(10-x)+sqrt(x^2-9) ?
The domain of f(x)=sqrt(10-x)+sqrt(x^2-9) is x<=-3\lor 3<= x<= 10