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

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Answer 1 · 2017-10-10T15:43:06

#t!=+-4# or #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}#