What is the domain of f(x)=sqrt(t^2-16) ?

The domain of f(x)=sqrt(t^2-16) is t = 4

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domain of f(x)=sqrt(t^2-16)

domain

f (x) = t^{2} - 16

t \leq - 4 or t \geq 4

Interval Notation

(- \infty, - 4] \cup [4, \infty)

Solution steps

Find non-negative values for radicals:

t \leq - 4 or t \geq 4

The function domain

t \leq - 4 or t \geq 4

f (x) = 14 - 7 x

f (x) = 7 x^{2} + 5 x - 3

f (x) = \frac{3}{x ^{2} - x - 12}

x^{\frac{5}{7}} - 8

f (x) = \frac{x + 6}{x ^{2} - 2 x - 48}

What is the domain of f(x)=sqrt(t^2-16) ?
The domain of f(x)=sqrt(t^2-16) is t<=-4\lor t>= 4