A Cubic Function and a Straight Line with One-variable Calculus

[seniorhs9]

Hi. I posted this in the Homework question but after 152 views with no right answer, my question looks analytical and rigorous enough to be posted here.

Thank you...

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I am asking about part iv).

[PLAIN]http://img715.imageshack.us/img715/7977/113ivb.jpg

Attempt at a solution

In the given fact, I think $x^3 - x - m(x - a)$ distance from the cubic function to the line. So for every point in $(b, c)$, this would be negative.

I'm really not sure how to show $c = -2b$ so I just tried to play with some algebra...

At x = b... $b^3 - b - m(b - a) = 0$

At x = c... $c^3 - c - m(c - a) = 0$

So they're both equal to 0...


$b^3 - b - m(b - a) = c^3 - c - m(c - a)$

so $b^3 - b - mb = c^3 - c - mc$

so by part i) $b^3 - b - b(3b^2 - 1) = c^3 - c - c(3b^2 - 1)$

so $b^3 - b - 3b^3 + b = c^3 - c - 3b^2c + c$

so $-2b^3 = c^3 - 3b^2c$

so $b^2(3c - 2b) = c^3$

but this doesn't look useful...

Thank you.

 

[omkar13]

FOR PART 4:
You just expand RHS of give eq. in and equate coefficients of x^2 on both sides and you'll get it.

 

[seniorhs9]

Hi omkar13. Thanks so much for your answer.

Now I feel like an idiot...how did you see to expand the RHS of the given eq.?

Somehow I didn't see it...I thought there was something harder going on...

 

[hotvette]

Wow, so simple. I actually found it a different way, but much more complicated.

https://www.physicsforums.com/showthread.php?t=547277