What is the domain of f(x)=sin(1/(t^2+1)) ?

The domain of f(x)=sin(1/(t^2+1)) is -infinity <t<infinity

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domain of f(x)=sin(1/(t^2+1))

domain

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

- \infty < t < \infty

Interval Notation

(- \infty, \infty)

Solution steps

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

- \infty < t < \infty

y = x ln (x)

f (x) = \frac{3 x + 4}{6 - 6 x}

f (x) = y^{2} - x - 4 y + 3 = 0

y = \frac{1}{x - 4}

10 x - x^{2}

What is the domain of f(x)=sin(1/(t^2+1)) ?
The domain of f(x)=sin(1/(t^2+1)) is -infinity <t<infinity