Q:Determine (up to a constant multiplier) the polynomial with a maximum at (-1,1), a minimum (1,-1) and no other critical points
It would be great if someone could show how this is determined. My initial thoughts are it would be in the form of $x^3 + x^2 + x + c$ but how do we determine the exact form.
Thanks!
Start with $f(x) = ax^{3} + bx^{2} + cx + d$. Since you want this polynomial to have critical points at $x = \pm 1$, we require that $f'(\pm 1) = 0$. This yields the two equations \begin{align*} 3a + 2b + c & = 0\\ 3a - 2b + c & = 0. \end{align*} It is then obvious that $b = 0$ and $c = -3a$. From here, one can obtain another two equations from the requirements that $f(1) = -1$ and $f(-1) = 1$. It is a simple matter to solve for $a$ and $d$.
Also note that this is the only such polynomial with degree at most 3. Also, there are no such functions with degree less than $3$. To see this, we observe that the derivative in this case will have degree at most $2$. Since it must, by assumption, have at least two distinct critical points at $\pm 1$, this forces $f'(x)$ to have degree $2$
However, if we allow for higher degree polynomials, then there may be infinitely many solutions since the critical points at $\pm 1$ may have arbitrary orders (i.e. they could be degenerate). By the condition on the critical points of $f$, we have the general formula \begin{equation*} f'(x) = a_{n,k}(x-1)^{k}(x+1)^{n}, \quad 1 \leq n,k \end{equation*} and $a_{n,k}$ is a constant to be determined by the constraints \begin{equation*} f(1) = -1, \quad f(-1) = 1. \end{equation*}
By the fundamental theorem of Calculus, we can then write \begin{equation*} f(x) = \int_{-1}^{x}a_{n,k}(t-1)^{k}(t+1)^{n}\,dt + f(-1) = \int_{-1}^{x}a_{n,k}(t-1)^{k}(t+1)^{n}\,dt + 1. \end{equation*} Then we can determine $a_{n,k}$ from the equation \begin{equation*} -1 = f(1) = \int_{-1}^{1}a_{n,k}(t-1)^{k}(t+1)^{n}\,dt + 1. \end{equation*} Making the substitution $u = t+1$, we have \begin{equation*} -2 = \int_{0}^{2}a_{n,k}(u-2)^{k}u^{n}\,du. \end{equation*}
Now integrate by parts $k$ times to get rid of the $(u-2)^{k}$ factor, and we have \begin{equation*} -2 = a_{n,k}\left[ \frac{k!n!}{(n+k+1)!}(-1)^{k}u^{n+k+1}\right]_{u=0}^{2} = a_{n,k} \frac{k!n!}{(n+k+1)!}(-1)^{k}2^{n+k+1}. \end{equation*} Thus \begin{equation*} a_{n,k} = \frac{(n+k+1)!(-1)^{k+1}}{n!k!2^{n+k}}. \end{equation*}
So we seek polynomials of the form \begin{equation*} f(x) = \frac{(n+k+1)!(-1)^{k+1}}{n!k!2^{n+k}} \int_{-1}^{x}(t-1)^{k}(t+1)^{n}\,dt + 1, \quad n, k \geq 1. \end{equation*} However, there is one last thing that we must check, which is to ensure that such $f$ don't have possible saddle points at $x = \pm 1$. For this, we use the constraints \begin{align*} f'(x) > 0 & \quad \textrm{ for } x < -1\\ f'(x) < 0 & \quad \textrm{ for } x > 1. \end{align*} Note that \begin{equation*} \mathrm{sign}(f'(x)) = \left\{ \begin{array}{cc} \mathrm{sign}(a_{n,k})(-1)^{k}(-1)^{n} & x < -1\\ \mathrm{sign}(a_{n,k})(-1)^{k} & -1 < x < 1\\ \mathrm{sign}(a_{n,k}) & 1 < x \end{array}\right. \end{equation*} From this we deduce that $a_{n,k} > 0$ and that $n,k$ are both odd. Also note that $a_{n,k} > 0$ whenever $k$ is odd so we just need the second condition. So, all solutions to the problem have the form \begin{equation*} f(x) = \frac{(n+k+1)!(-1)^{k+1}}{n!k!2^{n+k}} \int_{-1}^{x}(t-1)^{k}(t+1)^{n}\,dt + 1, \quad n, k \geq 1, \quad n,k \textrm{ odd }. \end{equation*}