What is the domain of f(x)=sqrt(16-((\sqrt{x+1))^2)} ?

The domain of f(x)=sqrt(16-((\sqrt{x+1))^2)} is -1<= x<= 15

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-((\sqrt{x+1))^2)}

domain

f (x) = 16 - ((x + 1)^{2})

- 1 \leq x \leq 15

Interval Notation

[- 1, 15]

Solution steps

Find non-negative values for radicals:

- 1 \leq x \leq 15

The function domain

- 1 \leq x \leq 15

16 - x^{2}

f (x) = 4 x - 16

\frac{x + 2}{x ^{2} + 3 x - 10}

x^{2} + 2 x - 3

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

What is the domain of f(x)=sqrt(16-((\sqrt{x+1))^2)} ?
The domain of f(x)=sqrt(16-((\sqrt{x+1))^2)} is -1<= x<= 15