Prove that rational function has no vertical tangent lines

[PaulKraemer]

Hi,

I have a problem that asks me to prove that the graph of a rational function has no vertical tangent lines.

I assume that to do this, I have to prove that where f(x) is a rational function, there can be no number 'a' where the limit as x->a of |f '(x) | = infinity.

From my understanding, a rational function is simply one polynomial function divided by another. When you take a derivative using the quotient rule, you end up with with the denominator equal to the denominator of the original function squared. If this denominator has a 'zero', then I can't see why it would be impossible to find a number 'a' where the limit as x->a of |f '(x) | = infinity.

If anyone can clear this up for me, I'd really appreciate it.

Thanks in advance,
Paul

 

[JeffM]

Re: Prove that rational function has no vertical tangent lin

Hi,

I have a problem that asks me to prove that the graph of a rational function has no vertical tangent lines.

I assume that to do this, I have to prove that where f(x) is a rational function, there can be no number 'a' where the limit as x->a of |f '(x) | = infinity.

From my understanding, a rational function is simply one polynomial function divided by another. When you take a derivative using the quotient rule, you end up with with the denominator equal to the denominator of the original function squared. If this denominator has a 'zero', then I can't see why it would be impossible to find a number 'a' where the limit as x->a of |f '(x) | = infinity.

If anyone can clear this up for me, I'd really appreciate it.

Thanks in advance,
Paul

Hi Paul

I CANNOT clear it up because f(x) = (1/x) seems to me to be a rational function where f'(x) approaches plus and minus infinity as x approaches 0. Are there no other limitations imposed on the rational function in your problem?

It is quite possible that I am making a donkey of myself by demonstrating that I do not know the definition of a rational function. If so, galactus, subhotosh khan, or tkhunny will be along after a while to put you on the right track.

PS If it is a specific rational function, have you checked whether the denominator of that specific function does have any zeroes?

PPS If it is a specific rational function that has a denominator with one or more real zeroes, can they be factored out of both the numerator and denominator?

 

[PaulKraemer]

Re: Prove that rational function has no vertical tangent lin

Hi Jeff,

It is not a specific rational function and there are no other limitations imposed in the problem...and just in case I may have mis-stated it, I will repeat the exact wording from the book:

"Prove that the graph of a rational function has no vertical tangent lines."

Your argument using f(x) = 1/x makes sense to me...perhaps I ran into a mistake in the book or like you said, maybe I don't know the correct definition of rational function?

Thanks for your help,
Paul

 

[galactus]

Re: Prove that rational function has no vertical tangent lin

Some thoughts:

If a function has a vertical tangent, then the derivative becomes unbounded there and the value is in the domain of the function.

But, rational functions have vertical asymptotes at the values that make the denominator equal to 0.

Thus, the derivative does not exist at that point. The value is not in the domain of the function.

 

[JeffM]

Re: Prove that rational function has no vertical tangent lin

Thanks galactus for clearing that up.

Sorry Paul, but I told you someone would fix my errors.