What is the domain of f(x)=sqrt(16-x^2)+sqrt(3-x) ?

The domain of f(x)=sqrt(16-x^2)+sqrt(3-x) is -4<= x<= 3

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(16-x^2)+sqrt(3-x)

domain

f (x) = 16 - x^{2} + 3 - x

- 4 \leq x \leq 3

Interval Notation

[- 4, 3]

Solution steps

Find non-negative values for radicals:

- 4 \leq x \leq 3

The function domain

- 4 \leq x \leq 3

f (x) = \frac{x ^{2} - 3 x - 4}{- x + 5}

\frac{x}{x + 1}

y = \frac{1}{9 - x ^{2}}

f (x) = \frac{4}{3 - x}

- x^{2} + 1

What is the domain of f(x)=sqrt(16-x^2)+sqrt(3-x) ?
The domain of f(x)=sqrt(16-x^2)+sqrt(3-x) is -4<= x<= 3