|
J. asks,
Consider the function f(x)=2x3+6x2-4.5x -13.5. State the roots of this cubic and confirm using the remainder theorem. Then, taking the roots two at a time, find the equations of the tangent lines to the average of two of the three roots. Find where the tangent lines at the average of the two roots intersect the curve again. Does this observation hold regardless of which two roots you average? State a conjecture concerning the roots of the cubic and the tangent lines at the average value of these roots.
Proof:
Let the three roots of f(x) be m, n, and p. Then, for some non-zero a,
f(x) = a(x-m)(x-n)(x-p)
Without loss of generality, we will find the line tangent to f at the average of m and n. First, we will find the point P ((m+n)/2, f((m+n)/2)) in terms of m, n, and p:
(8/a) f(x) = 8(x-m)(x-n)(x-p)
(8/a) f((m+n)/2) = 8(((m+n)/2)-m)(((m+n)/2)-n)(((m+n)/2)-p)
= (-m+n)(m-n)(m+n-2p)
= (2p-m-n)(m-n)2
So point P is ((m+n)/2, (a/8)(2p-m-n)(m-n)2). Now, we will find the slope of the tangent line at point P. Differentiating f,
(8/a) f'(x) = 8((x-m)(x-n)+(x-m)(x-p)+(x-n)(x-p))
(8/a) f'(x) = 2((2x-2m)(2x-2n)+(2x-2m)(2x-2p)+(2x-2n)(2x-2p))
The value of f'(x) when x=(m+n)/2 is the slope of the tangent line at point P, so we substitute m+n in place of 2x:
(8/a) f'((m+n)/2) = 2((m+n-2m)(m+n-2n)+(m+n-2m)(m+n-2p)+(m+n-2n)(m+n-2p))
= 2((-m+n)(m-n)+(-m+n)(m+n-2p)+(m-n)(m+n-2p))
= 2((-m+n)(m-n)
= (-2)(m-n)2
The equation of a line with slope m that passes through point (x0,y0) is
y - y0 = m (x-x0)
So the tangent line at point P, which is ((m+n)/2, (a/8)(2p-m-n)(m-n)2), is given by
y - (a/8) (2p-m-n)(m-n)2 = (a/8)(-2)(m-n)2 (x-(m+n)/2)
y = (a/8)(-2)(m-n)2 (x-(m+n)/2) + (a/8)(2p-m-n)(m-n)2
y = (a/8)(-2)(m-n)2 (x-(m+n)/2-p+m/2+n/2)
y = (-a/4)(m-n)2 (x-p),
which is the equation of a line that goes through the point (p,0), proving the conjecture
|
Nrich, Zeros of Cubic Functions and Zeros of Cubic Functions
The Cubic Formula -- the general solution of a degree three polynomial
The webmaster and author of this Math Help site is Graeme McRae.