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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: HelpmehplZ on June 04, 2014, 11:31:27 pm

Title: MATHEMATICAL METHODS CAS QUESTION- calculus explanation
Post by: HelpmehplZ on June 04, 2014, 11:31:27 pm: Hi I was asked a really awkward question and I need an explanation of why the midpoint of 2 x intercepts of a cubic has a tangent that intercepts the same x intercept as the cubic. We needed to do this with different combinations of the x intercepts e.g x int 1 and 3. 3 and 2 etc. and for some odd reason the tangent equation of that point has the same x int as the cubic. ANy help would be nice thanks!
Title: Re: MATHEMATICAL METHODS CAS QUESTION- calculus explanation
Post by: Zealous on June 05, 2014, 04:35:05 pm: Awkward question? here's an awkward algebraic proof.
(using the CAS...)

1. Start by setting up an equation for f(x) for a cubic in intercept form. For this cubic, we have intercepts (a,0), (b,0) and (c,0):

$\begin{eqnarray*}\\let\ f(x) & = & (x-a)(x-b)(x-c)\\f'(x) & = & 3x^{2}+(-2a-2(b+c))x+a(b+c)+bc\\\end{eqnarray*}\\$

2. Use $y-y_1=m(x-x_1)$ to find the equation of the tangent line of this cubic function at x=p. Then let y=0 so we can solve for x, which will be the x intercept of the linear line.

$\begin{eqnarray*}\\Tangent\ line\ at\ x & = & p\ \ \ y=f(p)\\y-f(p) & = & f'(p)(x-p)\\\\Let\ y=0\ find\ x\ intercept:\\0-f(p) & = & f'(p)(x-p)\\-\frac{f(p)}{f'(p)} & = & (x-p)\\x_{int} & = & p-\frac{f(p)}{f'(p)}\\\end{eqnarray*}\\$

3. We have now found the x intercept of the tangent to f(x) at any point x=p. Now sub in the x coordinate of the point half way between intercepts (sub this into p):

$\begin{eqnarray*}\\Sub\ in\ p & = & \frac{b+c}{2}\ \ (half\ way\ between\ x\ intercepts)\\x_{int} & = & a\\\\Sub\ in\ p & = & \frac{a+c}{2}\\x_{int} & = & b\\\\Sub\ in\ p & = & \frac{a+b}{2}\\x_{int} & = & c\\\end{eqnarray*}\\$

So when we take the tangent line of f(x) where x is half way between the a and b, the x intercept of the tangent line will be at c. The same occurs for taking the midpoint of other intercepts.

Here's a graph for those who are interested. If we take the tangent line at x=1, which is half way between the (-1,0) and (3,0) intercepts, the tangent's x intercept will be at (-3,0) which is the other intercept of the cubic graph.

$f(x)=(x+1)(x-3)(x+3)$
(http://i1282.photobucket.com/albums/a531/Ovazealous/screenshot171_zpsb219fbcb.png)

Pretty interesting question, thanks for sharing!