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

The domain of f(x)=sqrt(t^2-9) is t = 3

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

domain

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

t \leq - 3 or t \geq 3

Interval Notation

(- \infty, - 3] \cup [3, \infty)

Solution steps

Find non-negative values for radicals:

t \leq - 3 or t \geq 3

The function domain

t \leq - 3 or t \geq 3

f (x) = \frac{1}{1 + x ^{2}}

y = - 2 x + 2

\frac{x ^{2} - 9}{x ^{2} + 1}

y = 2 - x

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

What is the domain of f(x)=sqrt(t^2-9) ?
The domain of f(x)=sqrt(t^2-9) is t<=-3\lor t>= 3