Quote (Forg0tten @ 18 Apr 2018 23:36)
Don't need it solved but perhaps it's a fun exercise for someone.
f(x) = x^3 - 11x^2 - 25x - 13
y = p*x + q
Line Y is the tangent of f(x) at A(a,f(a)) and intersects with f(x) at B (13,0). What is the sum of p+q?
So let's see, we have 13p + q = 0, so p + q = - 12p
Let's find p. Since y is the equation associated with the tangent of f at (a,f(a)), we have that p = f'(a), so all we need is to find a.
f(a) + (13-a)f'(a) = 0 <=> a^3 - 11a^2 - 25a - 13 + (13 - a)(3a^2 - 22a -25) = 0
We already know that 13 is one of the roots of the third-degree polynomial, and it is possible either to proceed by identification (boring) or to simply guess that -1 is also a root, so a^3 - 11a^2 - 25a - 13 + (13 - a)(3a^2 - 22a -25) = -2(a - 13)(a + 1)(a - c) with -2*13c = -13(25 + 1) = -13*2*13 -> c = 13
So a^3 - 11a^2 - 25a - 13 + (13 - a)(3a^2 - 22a -25) = -2(a+1)(a - 13)^2
The only possible solutions are either -1 or 13.
Now it depends if "intersect" means simply share a point or cut across. I think the latter is what is meant, but anyway : In both cases -1 works and p = f'(-1) = 3 + 22 - 25 = 0 so
p + q = 0. This is the only solution for the second case since at 13 f is convex (f''(13) = 6*13 - 22 = 78 - 22 = 56 > 0 so it does not "cut across" but stays under the curve
In the first case, there is also p = f'(13) = 3 * 169 -22 * 13 - 25 = 507 - 286 - 25 = 507 - 306 - 5 = 201 - 5 = 196 so
p + q = - 12 * 196 = - 2280 - 72 = -2352Geez, and this was supposed to be fun? I was already bored to tears and now this is the final nail in the coffin lmao
Oh and also,
Quote (jordando101 @ 10 Mar 2018 07:40)
So i just found this section of JSP, been on JSP for years. Also been a high school math teacher for years (undergrad in math, masters in computer science).
Willing to help and give good, detailed answers for fg!
Just shoot me a pm :)
Wait, you're
asking for fg when you're a high school teacher (i.e. failed maths major) and all you can solve are these boring exercises? Haha, are you Jewish?
This post was edited by Hanako on Jun 5 2018 02:02pm