Another calc prob if you don't mind

[martindcx1e]

Alright here's one more for yall. I don't even know where to start on this kind of problem. Help plz?

Show that the equation x^101 + x^51 + x - 1 = 0 has exactly one real root.

[mcatdog]

hint #1: graph its derivative
hint #2: evaluate f(0) and f(1)

[martindcx1e]

what if i'm supposed to do it without graphing? also, i'd like the most general method of showing this so i can do it for different equations too.

[mcatdog]

I'm trying to let you figure it out without giving it away

just take the derivative and see if you can figure out any properties that it might have, if not then try to graph it

[spoonitnow]

OP, You're seriously overthinking this imo.

[Miffed22001]

ill be damned if i know the answer

[gabe]

I'm trying to let you figure it out without giving it away

lame!!!

[spoonitnow]

This is how bored I am.

Let f(x) = x^101 + x^51 + x − 1.

First we prove existence of some number of roots: Since the limit as x approaches -infinity is -infinity and the limit as x approaches infinity is infinity, there has to be at least one root on (-infinity, infinity).

Now we narrow it down: When we take the derivative, we get f'(x) = 101x^100 + 51x^50 + 1. We note that f'(x) > 0 for all x (all powers are even, all coefficients are positive, and the constant term is positive). Now if there was more than one root, then by Rolle's Theorem, there would be a c such that f'(c) = 0. However, f'(c) is always positive, so there is only one root.

Hit me back in this thread or a PM if you don't get it. Applied Math major ftw.

[Miffed22001]

im still stuggling with E=mc2 ffs.