Graphs of Logarithmic Functions

[eutas1]

I don't understand the question or the worked solution...

 

[Dr.Peterson]

I don't understand the question or the worked solution...

They are asking how to restrict the function's domain so that its inverse is a function -- that is, so that the restricted function is one-to-one.

It will not be one-to-one if there are two values of x in the domain that yield the same value for f(x). That will happen if some value x and its negative -x are both in the interval. To prevent this, a must be positive, and the only choice in the list that ensures that is (b). In fact, as long as b > 0 it would work.

 

[Steven G]

I would look at it slightly differently.

f(x) = log (x⁴) = 4log |x|. It should be clearis clear that for example, f(-7) = f(7) which prevents f(x) from being 1-1 unless we restrict the domain. Since it must be (a, inf), we must have a>0. This is true only for choice b.

 

[eutas1]

They are asking how to restrict the function's domain so that its inverse is a function -- that is, so that the restricted function is one-to-one.

It will not be one-to-one if there are two values of x in the domain that yield the same value for f(x). That will happen if some value x and its negative -x are both in the interval. To prevent this, a must be positive, and the only choice in the list that ensures that is (b). In fact, as long as b > 0 it would work.

I can understand that the domain cannot contain negative values, because this would then produce TWO y-values --> therefore it is NOT a one-to-one function which means NO inverse function. But doesn't that mean 'a' should then be any positive number, therefore a > 0 ??? So can't the answer be (a), (b), and (d) since they are all positive?

 

[eutas1]

I would look at it slightly differently.

f(x) = log (x⁴) = 4log |x|. It should be clearis clear that for example, f(-7) = f(7) which prevents f(x) from being 1-1 unless we restrict the domain. Since it must be (a, inf), we must have a>0. This is true only for choice b.

If you do f(x) = log (x^4) =4log(x) , then wouldn't that mean x > 0 ? So then the answer would be (A), (B), and (D) ?

 

[Dr.Peterson]

I can understand that the domain cannot contain negative values, because this would then produce TWO y-values --> therefore it is NOT a one-to-one function which means NO inverse function. But doesn't that mean 'a' should then be any positive number, therefore a > 0 ??? So can't the answer be (a), (b), and (d) since they are all positive?

Read the problem very carefully. It says this:



It is not asking for the largest possible domain over which the function is one-to-one; it is asking WHICH of the four options will always lead to A possible domain. Also, it is not saying that \(a>1\) is the domain; it is saying that \([a,\infty)\) is a domain over which f would be one-to-one, for any value of a greater than 1. Yes, a can be any positive number; but that is not what they are saying.

If \(a\le 1\), that allows a to be negative, e.g. \(a = -1\), in which case the domain would be \([-1,\infty)\), and f would not be one-to-one.

The same is true for \(a\ge -1\), and for \(a<2\).

 

[eutas1]

Read the problem very carefully. It says this:

View attachment 26338

It is not asking for the largest possible domain over which the function is one-to-one; it is asking WHICH of the four options will always lead to A possible domain. Also, it is not saying that \(a>1\) is the domain; it is saying that \([a,\infty)\) is a domain over which f would be one-to-one, for any value of a greater than 1. Yes, a can be any positive number; but that is not what they are saying.

If \(a\le 1\), that allows a to be negative, e.g. \(a = -1\), in which case the domain would be \([-1,\infty)\), and f would not be one-to-one.

The same is true for \(a\ge -1\), and for \(a<2\).

OH! I misread the '<' signs as '>' so that's why I was confused... my bad! Thank you very much