If $(f(x)-x)f''(x)>0$ , then $f(x)=f^{-1}(x)$ has no solution.

Question

If $f:\mathbb{R}\to\mathbb{R}$ is a double differentiable bijective function, then which of the following statements may be true ?

(A) $(f(x)-x)f''(x)\leq 0$ for all $x\in\mathbb{R}$

(B) $(f(x)-x)f''(x)\geq 0$ for all $x\in\mathbb{R}$

(C) If $(f(x)-x)f''(x)>0$ , then $f(x)=f^{-1}(x)$ has no solution.

(D) If $(f(x)-x)f''(x)>0$ , then $f(x)=f^{-1}(x)$ has at least one real solution.

My Attempt

Clearly if $f(x)=e^{-x}$ then both (A) and (B) are NOT true.

If I take $f(x)=e^x$ then (C) appears to be true.

But is there a general explanation

Answer 1

If $(f(x)-x)f''(x)>0$ for all $x \in \Bbb R$ then in particular $f(x) -x \ne 0$ for all $x \in \Bbb R$. So one of the following two must be true: Either $$ \forall x \in \Bbb R: f(x) > x \\ \implies \forall x \in \Bbb R: f(x) > x > f^{-1}(x) \, . $$ or $$ \forall x \in \Bbb R: f(x) < x \\ \implies \forall x \in \Bbb R: f(x) < x < f^{-1}(x) \, . $$

In any case, $f(x) = f^{-1}(x)$ does not have a solution.