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There exists a differentiable function $f\colon\mathbb{R}\to\mathbb{R}$ with the following property: the tangent at each point has infinitely many common points with the graph

/Edit: $f$ nonlinear/

For $\sin x^2$ we have this property at point $0$, but it's hard to imagine this could happen at every point... I think the statement is false, but no idea how to prove it.

larry01
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3 Answers3

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$e^x\sin(x)$ meets any straight line infinitely many times.

More generally, $f(x)g(x)$ where $f$ has a superlinear growth and $g$ is periodic and alternates.

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Lets consider the differentiable function $y=f(x)$ and suppose $a$ be any point of domain of $f.$ Then $(a,f(a))$ is a point oi graph of the function. Then $$\dfrac{df}{dx}=\dfrac{y-f(a)}{x-a}$$ $$y=(x-a)\dfrac{df}{dx}+f(a)$$is the equation of the tangent of graph at $x=a.$
According to your condition, we want to find a function $f$ such that, the differential equation $$f(x)=(x-a)\dfrac{df}{dx}+f(a)$$ has infinitly many solutions for all $a$ in the domain of $f.$
(Polynomial function does not satisfy this equation unless it is linear or a constant.)
There can be uncountably many such functions.
Best way of proving your result is give an example of a such function as Yves Daoust did.

Bumblebee
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  • That doesn't look like a differential equation to me (given that it takes $x$ as the variable and I see no derivative of the real number $x$ involved, whatever that might mean) - and it's quite misleading to say "...the differential equation has infinitely many solutions", since that usually implies the solutions are functions. It looks like an equation which, as a constant. (And it would be better to notate $f(x)=(x-a)f'(a)+f(a)$, to clarify where the derivative is taken) – Milo Brandt Feb 12 '15 at 01:32
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A constant one $x\mapsto c, c\in\Bbb R$.

CiaPan
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