domain of: y=x+5/square root: (x^2-1)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

domain of: y=x+5/square root: (x^2-1)

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[y=\frac{x+5}{\sqrt{x^2-1}}\]The domain of this function is the set of all x such that the function is well-defined. You need to ensure two things here. 1. the numerator cannot be 0 2. the radiacand (expression under the sqrt sign) cannot be negative (for real numbers). Let's look for x-values the function CAN'T take. This will be when\[\sqrt{x^2-1}=0\]by the first point, and when\[x^2-1<0\]by the second point. For the first,\[\sqrt{x^2-1}=0 \rightarrow x^2-1=0 \rightarrow x= \pm 1\]For the second,\[x^2-1<0 \rightarrow x^2<1 \rightarrow -1
The above 'D' means 'the set of all real numbers, except for those between -1 and 1 inclusive of -1 and 1'.
You're other equation is right, btw.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

*Your
Okay, thanks for letting me know..

Not the answer you are looking for?

Search for more explanations.

Ask your own question