Is there a function $f$ such that $f(x)<0, f'(x)<0,f''(x)>0 \,\forall x\in \mathbb R$ ?

I think there is, since it would be quite strange that there is not (and maybe hard to prove). After trying some logarithms, exponentials and polynomials I am out of ideas.

Definitely polynomials doesn't work: If $f(x)$ is a polynomial such that $f(x)<0 \forall x\mathbb \in \mathbb R$ then it is clear that the leading exponent must be even and the coefficient of it has to be negative. But then after derivating we violate our initial condition for $f'(x)$. Any ideas?

Note: Given the level level of the other questions in the book, the most likely functions that a simple example would use are logarithms and exponentials, trigonometric functions and their inverses, and polynomials. However, any example is welcome.


There is no such function. Since $f''>0$ it is strictly convex, and its graph is above the tangent at any point. Then $$ f(x)\ge f(0)+f'(0)\,x\quad\forall x\in\mathbb{R}. $$ Since $f'(0)<0$, $f(x)>0$ if $x<-|f(0)/f'(0)|$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.