How do you find the domain of $g (t) = \frac{t + 4}{t^{2} - 16}$ $g(t)=(t+4)/(t^2-16)$ ?

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Answer 1 · 2017-10-10T15:43:06

$t \neq \pm 4$ $t!=+-4$ or $t \in (- \infty, + \infty) - {\pm 4}$ $t in(-oo,+oo)-{+-4}$ or $t in RR-{+-4}$

The function is $g(x)=(t+4)/(t^2-16$

Its written in the form of a fraction. As we know the denominator of a fraction can never be $0$ , therefore we'll apply a condition to the function.

$t^2-16!=0$

$t^2!=16$

$t!=sqrt16$

Therefore

$t!=+-4$

So $t$ can be any number except $+-4$

If you want to write it in interval notation it can be written as:-

$t in(-oo,+oo)-{+-4}$

or

$tinRR-{+-4}$

How do you find the domain of g(t)=(t+4)/(t^2-16)g(t)=t+4t2−16g(t)=(t+4)/(t^2-16)?