What is the domain of f(t)=sqrt(t+1) ?

The domain of f(t)=sqrt(t+1) is t>=-1

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domain of f(t)=sqrt(t+1)

domain

f (t) = t + 1

t \geq - 1

Interval Notation

[- 1, \infty)

Solution steps

Find non-negative values for radicals:

t \geq - 1

The function domain

t \geq - 1

y = 5 x + 5

\frac{3 x ^{2}}{x ^{2} + 1}

4 x + 1

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

f (x) = \frac{2 x - 1}{x - 3}