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Tangent Lines and Odd Degree Polynomials

Date: 07/24/98 at 16:13:42
From: Michelle Moulton
Subject: modern algebra/geometery

Let p(x) be a polynomial of odd degree. Determine whether every point 
in the plane lies on at least one line which is tangent to the curve  
y = p(x).

Date: 07/26/98 at 15:27:56
From: Doctor Jaffee
Subject: Re: modern algebra/geometery

Hi Michelle,

This was a tough one, but I think I've found a way to justify that 
every point does, in fact, lie on at least one line that is tangent to 
p(x), where p(x) is an odd polynomial. I assume that you understand the 
basics of calculus. If not, this explanation probably won't make any 
sense to you. In that case, write back and I'll try to come up with a 
non-calculus explanation.

Since the function is differentiable everywhere, any point on the 
curve is automatically on a line tangent to the curve. Let's consider 
the point (a, b), which is not on the curve. In order for this point to 
be on a line tangent to the curve, there must be a point (x, p(x)) 
which is the point of tangency. The slope of the line, then, is p'(x) 
and the equation of the line is p(x) - b = p'(x)(x - a), where a and b 
are constants and x is the variable we have to calculate to find the 
point of tangency.

In other words we have to solve an equation whose degree is odd. But 
the curve of every polynomial with an odd degree has to intercept the 
x-axis at least once, so this equation must have at least one solution, 
which proves that there does exist at least one point on the curve 
whose tangent line goes through the point (a, b).

I hope this explanation makes sense. If not, write back.

- Doctor Jaffee, The Math Forum
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Associated Topics:
College Calculus
College Euclidean Geometry

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