Invertible Functions: Solving Interval Problems

[mathgeek69]

1. Indicate whether the functions below are invertible within the given interval
2. a. sechx on [0, positive infinity)
2. b. e^(x^2) on (-1,2]
2. c. cos(lnx) on (0, e^pi]3. I attempted these but the intervals confuse me. Any special tricks you guys use? How do i solve these?

 

[jedishrfu]

Welcome to PF. This looks like a homework assignment. In order for us to help you, you need to follow the homework template and show some work first so we can point you in the right direction.

The first question is what is the definition of an invertible function?

 

[mathgeek69]

Alright. I solved e^(x^2) on (-1,2]. I got the following.

If F(x)= e^(x^2) on (-1,2]
then, F^-1(x)= sqrt(lny) , where F^-1(x) means the inverse function of F(x)

Is this correct? Now I do not understand what it means by on the given interval. Do i sub in the values of x=-1 and x=2 in the F^-1(x) function and see if an answer exist. If the answer exists for both, then the function F(x) is invertible on the given interval?

Is this the way of approaching this?

 

[Mark44]

Alright. I solved e^(x^2) on (-1,2]. I got the following.

What do you mean when you say you solved e^x2 on (-1, 2]? Do you mean you graphed it?

If F(x)= e^(x^2) on (-1,2]
then, F^-1(x)= sqrt(lny) , where F^-1(x) means the inverse function of F(x)

On one side you have a function of x, F^-1(x), but on the other side, you have something that involves only y. That's not right.

Is this correct? Now I do not understand what it means by on the given interval. Do i sub in the values of x=-1 and x=2 in the F^-1(x) function and see if an answer exist. If the answer exists for both, then the function F(x) is invertible on the given interval?

Is this the way of approaching this?

No. The first thing to do with all three of these is to sketch a graph of the functions. Just seeing the graph, if it's reasonably accurate, should be enough to be able to say if they are invertible. Note that the questions DON'T ask you for the formula of the inverse.

You have probably learned some rule about when a function has an inverse and when it doesn't. I would take a look at that rule.

 

[mathgeek69]

y= e^(x^2)
lny=lne^(x^2)
lny=x^2
x=sqrt(lny)

Interchage x&y

y=sqrt(lnx)

F^-1(x)= sqrt(lnx)

That is the inverted function of e^(x^2). So I know that yes the function is invertible. What I don't know is how to figure out if the function is invertable over the domain (-1,2). How do I figure that out ?

 

[mathgeek69]

Figured it out!

domain of F^-1(x) is: x>= 1 so therefore it is not invertable over the interval (-1,2]

 

[SammyS]

y= e^(x^2)
lny=lne^(x^2)
lny=x^2
x=sqrt(lny)

Interchage x&y

y=sqrt(lnx)

F^-1(x)= sqrt(lnx)

That is the inverted function of e^(x^2). So I know that yes the function is invertible. What I don't know is how to figure out if the function is invertable over the domain (-1,2). How do I figure that out ?

Don't forget, when solving an equation such as u² = a , you get u = ±√(a). The solution has a ± sign.

So solving ln(y) = x² for x gives x = ± √( ln(y) ) .

Figured it out!

domain of F^-1(x) is: x>= 1 so therefore it is not invertable over the interval (-1,2]

You may be right about this not being invertable, but not for the right reason.

 

[mathgeek69]

You may be right about this not being invertable, but not for the right reason.

What is the real reason behind the correct answer? I thought my reason was correct?

 

[haruspex]

What is the real reason behind the correct answer? I thought my reason was correct?

Your reasoning was wrong because you took as domain for F^-1 the domain of F. When discussing whether some function F is invertible over a domain, you should take the domain of the supposed inverse function as the range of F (over F's given domain). It follows that to determine invertibility you only need to ask whether there are two values x and y in the domain of F for which F(x) = F(y). If so, it is not invertible; if not, then given any z in the range of F there must be a unique x in its domain s.t. z = F(x).