If $f$ is continuous and differentiable, $f(0) = f'(0) = f'(1) = 0$, is there a tangent line to $f$ through the origin? I think I can show this is true if $f''$ is continuous by reducing to a different problem, but it feels like there should be a way do this directly and I can't find it.
EDIT: The point of tangency should be in $(0,1).$
 A: Note that finding the point of tangency $x_0$ is asking that $x_0 \in (0,1)$ satisfy $\frac{f(x_0)}{x_0} = f'(x_0) \iff f(x_0) = xf'(x_0) \iff f(x_0) - xf'(x_0) =0$. This immediately reminded me of the product rule $[xf(x)]' = f(x) + xf'(x)$, but we have a plus where there is supposed to be a minus. No worries, the quotient rule has a minus! And indeed $[\frac{f(x)}{x}]' =\frac{xf'(x)-f(x)}{x^2}$ does have the correct distribution of signs. Defining $g(x) = \frac{f(x)}{x}$ and then extending by continuity to $g(0) = \lim_{x\to 0} \frac{f(x)}{x} = f'(0)= 0$. It would be nice if $g(1)=f(1)=0$ because then we could apply Rolle's theorem and be done, but this is not true in general.
But we don't need Rolle's theorem to tell us that $g'(\xi)=0$. We have another theorem, the critical point theorem, that tells us that the derivative of a differentiable function at a maximum (say at $\xi$) must be $0$. Since $g(x)$ is a continuous function on a compact (i.e. closed and bounded) set $[0,1]$, it attains a maximum at $\xi\in[0,1]$. We would be done, except we require the point of tangency to be in $(0,1)$. So we just need to rule out the possibility that $\xi = 0$ or $1$.
The only remaining case is $f(1)\neq 0$, so suppose without loss of generality that $f(1)>0$ (if it's negative, just multiply $f$ by $-1$). This rules out that $\xi=0$ since $g(0) = 0 < f(1) = g(1)$ (again we define $\xi$ to be the point at which $g(x)$ is maximized on $[0,1]$). Because $f'(0)=f'(1)=0$, I claim that $f(x)$ must be strictly below the line $f(1)x$ near $x=0$ and strictly above that line near $x=1$ (I'll leave this to you to work through). That means that  $g(x)=\frac{f(x)}{x}>f(1)=g(1)$ for $x \in (1-\delta, 1)$, so indeed $g(1)$ is NOT a maximum of $g$ on $[0,1]$, and we are done.
Intuition: the fact that this proof hinged on finding a point $x_0$ such that $g(x) = \frac{f(x)}{x}$ has $0$ derivative can be pictured geometrically as follows: the formula $\frac{f(x)}{x}$ is the slope of the secant line from the origin to the point $(x,f(x))$, and at a point of tangency like in the following picture
,
the slope of the secant line increases until the point of tangency and then decreases. So it makes sense intuitively that the point of tangency is a critical point of the secant-line-slope function.
Lastly, I remark that while thinking of this answer, I came across an easier version of this problem $f$ continous in $[a,b]$, differentiable in $(a,b)$ (where $b>a>0$) such that $f(a)/a=f(b)/b$ by searching using the search engine https://approach0.xyz/search/ (which I recommend all MSE users use and benefit from).
