Vanishing right-derived functor implies right-exact

Question

Let $\mathbf{f}\colon \mathbf{A}\rightarrow \mathbf{B}$ be an additive functor between Abelian categories (not necessarily left-exact), $\mathbf{A}$ with enough injectives. Suppose that $\operatorname{RDer}^m\mathbf{f}\cong 0$ for all $m\geq 1$. Is it necessarily the case that $\mathbf{f}$ is right-exact? If not, what additional assumptions does one need to make?

Answer 1

Assuming that by $\text{RDer}^n\mathbf{f}$ you mean the functor obtained by applying $\mathbf{f}$ to an injective resolution and taking degree $n$ cohomology, then the answer is no.

For example, take $\mathbf{f}$ to be a non-zero functor that vanishes on injectives, such as $\text{Ext}^1(X,-)$ for some object $X$.