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I know that the derivative of an even degree polynomial is odd, hence the derivative must have at least one real solution. But, this real solution could correspond to an inflection point as well. So, how do I prove in general that an even degree polynomial must have at least one turning point ?

Edit : I don't know if a constant function counts as an even degree polynomial, but regardless let us exclude that case.

Edit 2: A turning point is a point where the derivative changes sign. A turning point must also be a local maximum or local minimum.

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  • $\begingroup$ Welcome to Math.SE! In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. $\endgroup$ – emma Jul 21 '17 at 3:27
  • $\begingroup$ Does a constant function count as an even degree polynomial? Other that that, every even function has a turning point at $x=0$ $\endgroup$ – WW1 Jul 21 '17 at 3:43
  • $\begingroup$ @WW1 But OP asks for even degree polynomials, not even function. $\endgroup$ – Brian Cheung Jul 21 '17 at 3:53
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The polynomial function is continuous. For $f(x)=ax^{2n}+p_{2n-1}(x)$, it holds true: $$\lim_\limits{x\to\pm \infty} f(x)=\begin{cases} +\infty, \ if \ a>0 \\ -\infty, \ if \ a<0\end{cases}.$$ Noting the above, there must be a global extreme point. P.S. Using similar logic one can show that an odd degree function has local extreme (non-inflection) points.

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If $p(x)$ is an even degree polynomial, degree at least 2 say, then $q(x)=p'(x)$ is of odd degree. Odd degree polynomials take on both positive and negative values, so there must be a point $c$ where $q(x)<0$ for $x<c$ and $q(x)>0$ for $x>c,$ or the reverse of this, for $x$ sufficiently near $c.$. Thus $c$ is a turning point.

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  • $\begingroup$ Forgive me if I'm mistaken, but doesn't this just show that there is a stationary point. I'm aware that an even degree polynomial must have a stationary point, but I want to show that that there exists a turning point(not all stationary points are turning points). $\endgroup$ – John Mack Jul 21 '17 at 3:50
  • $\begingroup$ @JohnMack you should probably say the definition of turning point in your question. I have never seen that phrase used in calculus. $\endgroup$ – Will Jagy Jul 21 '17 at 3:58
  • $\begingroup$ Basically, a turning point is a point at which the derivative changes sign.A turning point may be either a local maximum or local minimum. I have seen this phrase used sometimes, since I also saw this same phrase used in Wikipedia I thought it would be more well known. $\endgroup$ – John Mack Jul 21 '17 at 4:03
  • $\begingroup$ @JohnMack: This answer shows exactly that the derivative changes sign, namely at the point $x = c$. $\endgroup$ – Jesse Madnick Jul 21 '17 at 4:09
  • $\begingroup$ I apologize, I phrased my last comment incorrectly. A turning point is a point where the derivative changes sign, but it must also be a local maximum or minimum. $\endgroup$ – John Mack Jul 21 '17 at 4:16
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You don't say your background, but you probably know enough to show that, given that we demand the coefficient of $x^{2n}$ positive, that there is a fixed positive $N > 0$ such that your polynomial satisfies $f(x) > f(0)$ for $|x| > N.$

Once you have that, it follows that the point with $-N < x < N$ where the polynomial achieves its minimum is also the global minimum, which is more than you need to finish the result. There is at least one local minimum, that being the global minimum I have indicated.

As an example, $x^4 + 1$ has just the local minimum at $x=0.$ Notice that the second derivative is equal to zero at this point, even though it is a local minimum.

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