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.
Conjecture: The tangent to the cubic f(x) at the average of two roots
intersects f(x) at the third root.
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
