The *reciprocal* of a one-to-one function

[sl2382]

Homework Statement

If g(x) is one-to-one and differentiable at every point in R, then its reciprocal 1/g(x) is also differentiable at every point in its domain. True of False?

Homework Equations

The Attempt at a Solution

I have tried several times, and the statement seems to be right. I think 1/g(x) can be regarded as 1/u and u=g(x), so 1/g(x) ' = 1/u ' * g'(x)=g'(x)/g(x) (I don't know if this is right...), but I'm stucked here and can't go any further because I don't know how to prove a function to be one-to-one..

I'll be very grateful if anyone can help me.. Thanks!

 

[Dick]

(1/g(x))'=g'(x)/g(x) is wrong. Can you correct it? The only thing I can think of that would involve 'one-to-one' is if you have stated the problem wrong. Did it ask if the INVERSE FUNCTION g^(-1)(x) is differentiable, not the reciprocal (g(x))^(-1) or 1/g(x)?

 

[sl2382]

Hi, thank you so much. I calculated once again: g^2(x)'/g^2(x) Is this one right?

It is the reciprocal, the 1/g(x). The question asks for if 1/g(x) is differentiable at every point in its domain.

 

[Dick]

Hi, thank you so much. I calculated once again: g^2(x)'/g^2(x) Is this one right?

It is the reciprocal, the 1/g(x). The question asks for if 1/g(x) is differentiable at every point in its domain.

No, you haven't got it right yet. Use the quotient rule.

 

[sl2382]

No, you haven't got it right yet. Use the quotient rule.

Is it -g'(x)/g^2(x) ?

 

[Dick]

Is it -g'(x)/g^2(x) ?

Yes it is! Now you know g'(x) exists for all x. So the quotient -g'(x)/g(x)^2 exists everywhere that g(x) is not equal to zero, right? Are points where g(x)=0 in the domain of 1/g(x)? I'm still not seeing what this has to do with g(x) being one-to-one.

 

[Bacle2]

But what happens at points where f is 0?

 

[Dick]

But what happens at points where f is 0?

What's f??

 

[Bacle2]

Oh, I meant g.

 

[sl2382]

Yes it is! Now you know g'(x) exists for all x. So the quotient -g'(x)/g(x)^2 exists everywhere that g(x) is not equal to zero, right? Are points where g(x)=0 in the domain of 1/g(x)? I'm still not seeing what this has to do with g(x) being one-to-one.

Does one-to-one means that g(x) can't be 0? Since if g=0 then it can't pass the horizontal line test.

 

[Bacle2]

Does one-to-one means that g(x) can't be 0? Since if g=0 then it can't pass the horizontal line test.

g(x)=x is 1-1.

 

[Dick]

Does one-to-one means that g(x) can't be 0? Since if g=0 then it can't pass the horizontal line test.

Not at all. g(x)=x is one-to-one. And g(0)=0. As I said, I don't get what one-to-one has to do with this problem if it's reciprocal.

 

[sl2382]

Not at all. g(x)=x is one-to-one. And g(0)=0. As I said, I don't get what one-to-one has to do with this problem if it's reciprocal.

Oh I got it. Thank you so much!

 

[HallsofIvy]

The constant function g(x)= 0 for all x is not one- to- one. Is that what you were thinking of?