restrictions on domain precalcluls quesiton

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Explain the restrictions on the domain of F(x)=log(x-5)+1 - if there are any. 

Explain the restrictions on the domain of    (the cube root of x-2) +3  -if there are any. 

Sep 2nd, 2015

Thank you for the opportunity to help you with your question!

1. To take the 'log' of a number, it must be greater than zero.  You can look at the graph to see that the 'log' graph is never a negative x-value, or you can try taking the 'log' of a negative number in your calculator, and it should come back undefined.  So you can set the inner part of the log greater than zero.

x-5>0

x>5

So, domain is: (5,infinity)


2. For all cubed roots, there are no restrictions to the domain.  It's a simple rule you can just memorize.  The reason this works is because you can multiply three negative numbers with each other to get a negative number, so the inner part of the cubed root can be negative.  The inner part of a square root cannot be negative though.

Domain: (-infinity, +infinity)

Please let me know if you need any clarification. I'm always happy to answer your questions.
Sep 2nd, 2015

in addition, for F(x)=(e^x+4) +7, is there a value that would cause the function to be undefined? Are there restrictions of the domain/explain- if not explain why. 



Also, for both of the first 2, are there any values of x that would cause the function to be undefined? 

Sep 2nd, 2015

Domain for e^x is all real numbers, so (x+4) can equal anything.  No restrictions on the domain.  You can also see from the graph that the domain of x is (-infinity,+infinity).  The range, however, is limited.

For the first two, there are no other values of x that would cause the function to be undefined.

Do you need ranges as well, or just domains?

Sep 2nd, 2015

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