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Find the domain of r(t)?

Find the domain of r(t)=<sqrt(t^2), 1/(t-1), ln(t+1)>.

Answer: (-1, 1)U(1, positive infinity)

But how do you figure this out?

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2 Answers

The domain of sqrt(t^2) is all real numbers;

The domain of 1/(t-1) is t ≠ 1

The domain of ln(t+1) is t > -1

The conjunction of the above three domains is (-1, 1) U (1, +oo) <==Answer

Hello Sun!

"Domain" is what "t's" you can use.

To get started, try to "visualize" each function ...

*** sqrt(t^2) can take ANY "t";

*** 1/(t-1) "blows up" when t=1 -- a vertical asymptote; 

*** ln(t+1) means t>-1, because you cannot take ln(0) or a negative #.

Thus, you can use -1<t<1, and t>1 matching your indicated answer.

RECAP: 1/(t-1) blows up at t=1, but you can use any other "t",

and "ln(t+1)" can work only on a +#, so (t+1)>0, or t>-1. Best wishes ...