# Don't understand the fundamentals of this problem using MVT

1. Nov 23, 2016

### Arnoldjavs3

1. The problem statement, all variables and given/known data
http://prntscr.com/daze68

What I don't understand:

1. "P be a polynomial with degree n"
do these equations satisfy this description?:
$$p(x) = (x^2 + x)^n$$
$$p(x) = (5x^2 + 2x)^n$$
etc.
2. "C1 is a curve defined by y=p(x)"
c1 is essentially just the curve of the polynomial?

3. p(x) = e^x
This is meant to find points of intersection(which are also called roots? Correct me if i'm wrong) right? That there are n solutions to this?

4. h(x) = p(x) - e^x
Where does this h(x) function come from? I thought this would equal to 0 if you rearrange the equation from
3. Is it just a function introduced for the sake of answering this problem?

5. "Since p is of degree n, the nth derivative is just a constant"
What is this saying? That after differentiation n is still a constant? Since it's a polynomial it has to be some constant to begin wtih right?

6. How do they go from h(x) = p(x) - e^x to h^(n)(x) = 0 and why does it have only one solution/root????
If you differentiate then it becomes h'(x) = p'(x) - e^x right?

I know that the MVT states( i think ) that the derivative will have n-1 roots. There exists a root between the interval of its original function but how does that apply here? Sorry if this is inappropiate for a question i just don't understand the problem at its core.

2. Nov 23, 2016

### Staff: Mentor

No. They are of degree $2n$.
Yes, or better: the graph $\{ (x,p(x)) \, \vert \, x \in \mathbb{R}\}$ of the polynomial.
Yes. The term root means "solution of the equation". And at most $n$. There might be less.
Just a definition. It is a "help" function. Instead of looking for solutions of $p(x)=e^x$ we now may look for zeroes of $h(x)$.
So, the answer is yes. $e^x$ looks like a bent $x-$axis. So the goal is to show, that what applies to the flat version also applies to the curved version.
I don't understand what's your issue here. $x^n$ becomes $const. \, x^1$ after $n-1$ differentiations, $const.$ after $n$ and $0$ after $n+1$ differentiations. $n$ as the exponent is reduced by $1$ with every differentiation. But the (constant) coefficients of the highest term changes, however only from one constant to another, e.g. $2x^n \rightarrow 2nx^{n-1} \rightarrow 2n(n-1)x^{n-2} \rightarrow \ldots$
Yes, and if you do it $n$ times, then it reads $h^{(n)}(x)=const. \, - \, e^x$ which has only one point, where $h^{(n)}=0$ is possible. (Why?)
The trick is to go upwards again from $h^{(n)}=0$ since $h^{n}(x) = h'\, (h^{(n-1)}(x) = 0$ and climb back to $h(x)$ of which we want to find at most $n$ zeroes (roots). But I haven't read nor done the proof, so this is to be taken with a grain of salt.