About the domain of points having tangent to a curve

Question

Let the graph of $y=f(x)$ be a curve $C$ and $f''>0$.

Prove that if $y_0\leq f(x_0)$ then there exist a tangent of $C$ go through $(x_0,y_0)$

I don't know how to prove the existence.

Answer 1

Hmm, as far as I can see, the claim is simply wrong. Here's a counterexample:

As idm shows in his answer, a necessary criterion for the claim to hold for a given function $f$, fixed $x_0$ and all $y_0$ is that the function $$g(x) = f(x) - (x-x_0)f'(x)$$ has image $\mathrm{im}(g) = (-\infty,g(x_0)]$, so let us come up with a function $f$ that does not have that property. One finds that $g'(x) = (x-x_0)f''(x)$, so it's clear that a counterexample would have to have $f''(x)$ approach $0$ rather quickly at $\pm \infty$, or in other words, $f'(x)$ needs to be bounded and approach its limits rather quickly, so here's something constructed to satisfy that:

Concretely, let $f(x) = \log(1+e^x)$. Then $f'(x) = e^x/(1+e^x)$, and $$f''(x) = \frac{e^x}{1+e^x} - \frac{e^{2x}}{(1+e^x)^2} = (1-f'(x))f'(x).$$ I'll let you verify that the latter function is positive.

Now, letting $x_0 = 0$, you can also verify that $\lim_{x \to \pm \infty} g(x) = 0$ (and $g(x) > 0$), so that as long as $y_0 \leq 0$, you won't be able to find the desired tangent.

It might be helpful to actually plot this function to visualize what's going on, so here's a plot of $f$ as well as one with 200 tangents.

Answer 2

Let $x_0\in\mathbb R$. We have to show that $$\exists t\in\mathbb R: y_0=f'(t)(x_0-t)+f(t)$$

I supposed that $f\in\mathcal C^2(\mathbb R)$, because if not, $f''$ wouldn't have any sense.

By hypothesis, $f''>0$ therefore $f$ is convex, and thus, $$\forall y\in\mathbb R, \ f(x_0)\geq f'(t)(x_0-t)+f(t).$$

Let consider $g:\mathbb R\to\mathbb R$ define by $$g(t)=f'(t)(x_0-t)+f(t).$$

We can remarque that $g\in\mathcal C^1(\mathbb R )$. Remark that $g(x_0)=f(x_0)$ and so $g(t)\leq g(x_0)$ for all $t\in\mathbb R$ (in other word, $g(x_0)$ is the maximum of $g$). Moreover $$g(\mathbb R)=]-\infty ,g(x_0)].$$

Therefore, if $y_0\leq f(x_0)=g(x_0)$, there is a $t\in\mathbb R$ such that $$y_0=g(t)=f'(t)(x_0-t)+f(t).$$

Q.E.D.

Answer 3

We can assume without loss of genrality that $y_0<f(x_0)$.

$f''>0$ implies that your function is strictly convex, therefore, once the point $(x_0,y_0)$ is "below" the graph of $f$, it is possible to find a linear function $$g_{a}(x)=ax-ax_0+y_0,$$

such that $$g_{a}(x_0)=y_0, \ f(x)> g_{a}(x),\ \forall \ x\tag{1}.$$

Now, find $a,x'$ such that $f(x')=g_a(x')$ and $f'(x')=a$ (i.e. rotate the line around the point $(x_0,y_0)$). This is possible, because of $(1)$.