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?
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?
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 ...