What is the domain of f(x)=sqrt(t^2+9) ?

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

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domain of f(x)=sqrt(t^2+9)

domain

f (x) = t^{2} + 9

- \infty < t < \infty

Solution steps

Find non-negative values for radicals:

- \infty < t < \infty

The function has no undefined points nor domain constraints. Therefore, the domain is

- \infty < t < \infty

\frac{x ^{2} - 6 x + 12}{x - 4}

f (x) = 3 x^{3} - 36 x - 6

f (x) = sin^{2} (2 x)

f (x) = x - 4

f (x) = 6 - 5 x^{2}

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