Affine to linear like conversion of a concave function

Question

Is the following true:

$$\log \left( \frac{1}{f(x)+K}\right)\mathrm{is\;concave}\Longleftrightarrow \log \left( \frac{1}{f(x)}\right)\mathrm{is\;concave},$$ where $K\in\mathbb{R} $ and $f(x),f(x)+K>0$?$

Answer 1

No. Consider a smooth function $f$ and let $g(x)=\log(1/f(x))=-\log(f(x))$. Differentiate to get $g'(x)=-f'(x)/f(x)$, and repeat to get $$g''(x)=\frac{\bigl(f'(x)\bigr)^2-f(x)f''(x)}{\bigl(f(x)\bigr)^2}$$ so $g$ is (strictly) concave iff $$\bigl(f'(x)\bigr)^2<f(x)f''(x).$$ If you were to replace $f(x)$ by $f(x)+K$ this inequality becomes instead $$\bigl(f'(x)\bigr)^2<\bigl(f(x)+K\bigr)f''(x),$$ and clearly the two inequalities are not equivalent.