Proof that every point can lie on a tangent to a curve

Question

for which odd degree polynomials will every point in the plane lie on at least one tangent to the curve p(x)? What if P(x) is even?

Answer 1

Answer: For all odd degree polynomials of degree $\ge 3$.

If $p$ is of degree $1$, then the graph of $p$ is itself the tangent line to all its points. Hence for any other point there is no tangent passing through.

Let $p(x)$ be a polynomial of odd degree $\ge 3$. By switching to $-p(x)$ if necessary, we may assume that the leading term is positive. We shall use only the following properties of $p$, which apply to a much wider range of functions:

1. $p'(x)$ exists for all $x\in\mathbb R$
2. $\lim_{|x|\to\infty} \frac{p(x)}{x}=+\infty$

Let $(a,b)$ be a point in the plane. If $b=p(a)$, the point is on the graph of $p$ and hence on the tangent at this point. If $p(a)\ne b$, consider the function $$\begin{align}f\colon \mathbb R\setminus\{a\}&\to\mathbb R\\x\ &\mapsto \frac{p(x)-b}{x-a}\end{align}$$ (that is, the slope of the line through $(x,p(x))$ and $(a,b)$. Assume $x_0\in\mathbb R\setminus\{a\}$ is a critical point of $f$. Then $$ 0=f'(x_0)=\frac{p'(x_0)(x_0-a)-(p(x_0)-b)}{(x_0-a)^2}\implies p'(x_0)=\frac{p(x_0)-b}{x_0-a}=f(x_0),$$ i.e., the line through $(x_0,p(x_0))$ and $(a,b)$ has the same slope as the tangent and finally is the tangent at $(x_0,p(x_0))$. We see therefore that it suffices to show that $f$ has a critical point in its domain. For this note that $f(x)\to+\infty$ as $x\to\pm\infty$ by property 2 above. If $p(a)>b$ [resp. $p(a)<b$] we have $f(x)\to+\infty$ as $x\to a^-$ [resp. $x\to a^+$]. Hence the lemma below applies to $f|_{(-\infty,a)}$ [resp. $f|_{(a,\infty)}$] so that $f$ has somewhere in its domain a local minimum and (as $f$ is differentiable per property 1 above) a critical point as desired.

Lemma. Let $(u,v)\subseteq \mathbb R$ be an open interval (with possibly $u=-\infty$ and/or $v=+\infty$). Let $f\colon (u,v)\to \mathbb R$ be a continuous function with $\lim_{x\to u}f(x)=\lim_{x\to u}f(x)=+\infty$. Then $f$ has a global minimum in $(u,v)$.

Proof. Pick $w\in (u,v)$. There exists $u_1>u$ such that $f(x)>f(w)+1$ for all $x\in(u,u_1)$; and there exists $v_1<v$ such that $f(x)>f(w)+1$ for all $x\in(v_1,v)$. On the compact interval $[u_1,v_1]$ the continuous function $f|_{[u',v']}$ attains its minimum, in $\xi$ say. As certainly $e\in[u',v']$, we have $f(\xi)\le f(w)$ and conclude that $\xi$ is also global minimum for all of $f$. $_\square$

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If $\deg p$ is even, there may exist points that are on no tangent to $p$ at all. Just consider $p(x)=x^{2n}$. You will quickly verify that all tangents to this $p$ cross the $y$-axis on or below the $x$-axis. Thus the point $(0,1)$ is on no tangent. and the point $(0,1)$. These examples are not even special:

Proposition. For all polynomials of even degree there exist points in the plane that are not on any tangent to the graph of $p$.

Proof. The case of degree $0$ is trivial. We may assume wlog. that the leading term of $p$ is positive. Then there exists an interval $[u,v]$ such that $p'(x)\le 0$ and $p''(x)\ge 0$ for all $x<u$ and $p'(x)\ge 0$ and $p''(x)\ge 0$ for all $x>v$. Pick $a\in(u,v)$. Then the function $$\tilde p(x)=\begin{cases}p(x)&\text{if $x\le u$ or $x\ge v$}\\ \frac{x-u}{v-u}(f(v)-f(u))\end{cases}$$ is convex. Therefore, any tangent to a point $(x,p(x))$ with $x\notin [u,v]$ runs on or below the line segment introduced in $\tilde p$. Especially, $(a,b)$ lies above all these tangents as soon as $b>\tilde p(a)$. For tangents to points $(x,p(x))$ with $x\in [u,v]$ note that $p$ and $p'$ are bounded on the compact interval $[u,v]$, say $|p(x)|\le A$ , $|p'(x)|\le B$ for all $x\in [u,v]$. Then over the interval $[u,v]$, all these tangents stay below $A+(v-u)B$. We conclude that $(a,b)$ is not on any tangent to $p$ if we let $b>\max\{\tilde p(a),A+(v-u)B\}$. $_\square$

Remark. As in the odd case we did not use much of the polynomial-ness of $p$. It suffices that

• $p'(x)$ exists for all $x\in\mathbb R$ and $p'$ is continuous
• $p$ is convex outside a bounded interval

Answer 2

Let $P=(a,b)$ a point and $y=f(x)$ a derivable function $\mathbb{R} \rightarrow \mathbb{R}$. Obviously if $b=f(a)$ we have a tangent from $P$. Otherwise a tangent to the graph of the function form $P$ exists if there is a point $X=(x,f(x))$ such that. $$ \dfrac{f(x)-b}{x-a}=f'(x) $$ So there exists a tangent to $f(x)$ from any point $P$ if this equation have at least one real solution $\forall (a,b) \in \mathbb{R}\times \mathbb{R}$.

If $f(x)$ is a polynomial of degree $n>1$, this equation is a polynomial equation of degree $n$ and it has certainly a real solution only if $n$ is odd.