A question from a calc 1 student

  • Thread starter Thread starter 1MileCrash
  • Start date Start date
  • Tags Tags
    Calc 1 Student
1MileCrash
Messages
1,338
Reaction score
41
[Mentor's note: This post does not contain the template for thread-starting posts in the homework forums, because it was moved from another forum.]

"If f'(x) gives me the slope of f at some point, why doesn't f'(x)x + b = f(x)?"

I have no idea how to answer this. Can you?

How about: it is the slope of the tangent line to some point of the function, and that set of tangent lines do not share the same y-intercept b necessarily, so b is not constant.
 
Last edited by a moderator:
Physics news on Phys.org
What is b? ;)

and that set of tangent lines do not share the same y-intercept b necessarily
I think this is the main point. For straight lines, the formula works (if we set b=f(0)), for other curves it does not work in general.
 
mfb said:
What is b? ;)

I think this is the main point. For straight lines, the formula works (if we set b=f(0)), for other curves it does not work in general.

Well, I believe the student was just referring to the "y = mx + b" idea they learn for general line equations. I have to say it was a great question as most calc 1 students seem to just seem to "go through the motions" and that's why I wanted to share.

Clearly, the underlying reason is what we're talking about, but I'm looking for a good way to answer the question without inducing confusion.


EDIT: Errm.. why was this moved to homework help? This is clearly not an actual inquiry nor is it a textbook style question at all.
 
Last edited:
jtbell moved the thread as it is a homework question (or very similar to one).

Clearly, the underlying reason is what we're talking about, but I'm looking for a good way to answer the question without inducing confusion.
Give a counterexample?
 
I think explaining why the equation of a straight line happens to be y=mx+b would be a good point to start after that move on to the link between differentials and tangent...
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top