What is the domain of f(x)=sqrt(72-x^2+x) ?

The domain of f(x)=sqrt(72-x^2+x) is -8<= x<= 9

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domain of f(x)=sqrt(72-x^2+x)

domain

f (x) = 72 - x^{2} + x

- 8 \leq x \leq 9

Interval Notation

[- 8, 9]

Solution steps

Find non-negative values for radicals:

- 8 \leq x \leq 9

The function domain

- 8 \leq x \leq 9

f (x) = \frac{x - 3 x - 10}{x ^{2} - 16}

- 8 x (1 - x^{2})^{3}

f (x) = (x - 3) (x + 5)

\frac{x + 3}{5 x - 1}

f (x) = 3 - 2 e x - 4

What is the domain of f(x)=sqrt(72-x^2+x) ?
The domain of f(x)=sqrt(72-x^2+x) is -8<= x<= 9