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

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

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domain of f(x)=sqrt(4-x^2)+sqrt(x+1)

domain

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

- 1 \leq x \leq 2

Solution steps

Find non-negative values for radicals:

- 1 \leq x \leq 2

The function domain

- 1 \leq x \leq 2

\frac{4 x - 2}{3 x + 1}

f (x) = \frac{x}{x + 4}

f (x) = - 2 x^{2} - 6 x + 20

f (x) = 2 x^{3} - x^{2} + 2 x - 1

f (x) = 4 x^{3} - 5 x^{2} + 2 x - 1

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