What is the domain of f(x)=t^2-sqrt(t^2-4)+(1+2t) ?

The domain of f(x)=t^2-sqrt(t^2-4)+(1+2t) is t = 2

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domain of f(x)=t^2-sqrt(t^2-4)+(1+2t)

domain

f (x) = t^{2} - t^{2} - 4 + (1 + 2 t)

t \leq - 2 or t \geq 2

Interval Notation

(- \infty, - 2] \cup [2, \infty)

Solution steps

Find non-negative values for radicals:

t \leq - 2 or t \geq 2

The function domain

t \leq - 2 or t \geq 2

\frac{x ^{2} - 3 x}{x ^{2}}

f (x) = x - \frac{5}{3}

f (x) = \frac{3 x - 5}{x - 2}

3 x - 21

f (z) = \frac{z ^{2} + 16}{z ^{2} - 3 z - 4}

What is the domain of f(x)=t^2-sqrt(t^2-4)+(1+2t) ?
The domain of f(x)=t^2-sqrt(t^2-4)+(1+2t) is t<=-2\lor t>= 2