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

The domain of f(x)=sqrt(16-x^2)+sqrt(x+1) is -1<= x<= 4

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domain of f(x)=sqrt(16-x^2)+sqrt(x+1)

domain

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

- 1 \leq x \leq 4

Interval Notation

[- 1, 4]

Solution steps

Find non-negative values for radicals:

- 1 \leq x \leq 4

The function domain

- 1 \leq x \leq 4

f (x) = 5 x - 15

\frac{1}{2} x + 3

\frac{1}{2} x - \frac{13}{2}

(- \frac{2}{6}) \frac{2}{3}

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

What is the domain of f(x)=sqrt(16-x^2)+sqrt(x+1) ?
The domain of f(x)=sqrt(16-x^2)+sqrt(x+1) is -1<= x<= 4