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?
Marked as Best Answer
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 ...
Already have an account? Log in
By signing up, I agree to Wyzant’s terms of use and privacy policy.
To present the tutors that are the best fit for you, we’ll need your ZIP code.
Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.
Good news! It looks like you already have an account registered with the email address you provided.
Log in Forgot your password?It looks like this is your first time here. Welcome!
To present the tutors that are the best fit for you, we’ll need your ZIP code.
Please try again, our system had a problem processing your request.
