Smooth Logarithm at zero/one with special conditions
There are a lot of possible approximations.
In particular, the variant $$\varphi(x) = \begin{cases} \log(1+a(x-1+\varepsilon)^6+b(x-1+\varepsilon)^5+c(x-1+\varepsilon)^4),\text{ if }x\in(1-\varepsilon, 1 + \varepsilon)\\[4pt] V(x), \text{ otherwise} \end{cases}\tag1$$ $$\varphi'(x) = \begin{cases} \dfrac{6a(x-1+\varepsilon)^5+5b(x-1+\varepsilon)^4+4c(x-1+\varepsilon)^3}{1+a(x-1+\varepsilon)^6+b(x-1+\varepsilon)^5+c(x-1+\varepsilon)^4},\text{ if }x\in(1-\varepsilon, 1 + \varepsilon)\\[4pt] V'(x), \text{ otherwise} \end{cases}$$ $$\varphi''(x) = \begin{cases} \dfrac{30a(x-1+\varepsilon)^4+20b(x-1+\varepsilon)^3+12c(x-1+\varepsilon)^2}{1+a(x-1+\varepsilon)^6+b(x-1+\varepsilon)^5+c(x-1+\varepsilon)^4}-\\[4pt] \dfrac{(6a(x-1+\varepsilon)^5+5b(x-1+\varepsilon)^4+4c(x-1+\varepsilon)^3)^2}{(1+a(x-1+\varepsilon)^6+b(x-1+\varepsilon)^5+c(x-1+\varepsilon)^3)^4},\text{ if }x\in(1-\varepsilon, 1 + \varepsilon)\\[4pt] V''(x), \text{ otherwise} \end{cases}$$ provides conditions in the point $x=1-\varepsilon.$
Other conditions can be presented in the form of \begin{cases} \log(1+64a\varepsilon^6+32b\varepsilon^5+16c\varepsilon^4) = \log(1+\varepsilon)\\[4pt] \dfrac{192a\varepsilon^5+80b\varepsilon^4+32c\varepsilon^3}{1+64a\varepsilon^6+32b\varepsilon^5+8c\varepsilon^4} = \dfrac1{1+\varepsilon}\\[4pt] \dfrac{(480a\varepsilon^4+160b\varepsilon^3+48c\varepsilon^2))(1+64a\varepsilon^6+32b\varepsilon^5+16c\varepsilon^4)-(192a\varepsilon^5+80b\varepsilon^3+32c\varepsilon^2)^2}{(1+64a\varepsilon^6+32b\varepsilon^5+16c\varepsilon^4)^2}\\[4pt] \qquad =-\dfrac1{(1+\varepsilon)^2}, \end{cases} or \begin{cases} 64a\varepsilon^5+32b\varepsilon^4+16c\varepsilon^3 = 1\\[4pt] 192a\varepsilon^5+80b\varepsilon^4+32c\varepsilon^3 = 1\\[4pt] 480a\varepsilon^4+160b\varepsilon^3+48c\varepsilon^2= 0,\\ \end{cases} \begin{cases} a = \dfrac1{32\varepsilon^5}\\[4pt] b = -\dfrac3{16\varepsilon^4}\\[4pt] c = \dfrac5{16\varepsilon^3}. \tag2\end{cases} Therefore, the approximation has form $(1)$ with the coefficients $(2).$
For $\varepsilon = 0.1$ see also Wolfram Alpha.
Addition of 09.06.2018
To prove of the condition $2,$ let us consider the polynomial $$P_6(y) = ay^6+by^5+cy^4$$ with the coefficients $(2),$ i.e. $$P_6(y) = \dfrac1{32\varepsilon^5}y^6-\dfrac3{16\varepsilon^4}y^5 +\dfrac5{16\varepsilon^3}y^4 =\dfrac1{32\varepsilon^5}(y^6-6\varepsilon y^5+10\varepsilon^2y^4).$$ Then $$\varphi(x) = \begin{cases} \log(1+P_6(x-1+\varepsilon)),\text{ if }x\in(1-\varepsilon, 1 + \varepsilon)\\[4pt] V(x), \text{ otherwise}. \end{cases}$$ Easy to see that $$P_6(y) = \dfrac1{32\varepsilon^5}y^4\left((y-3\varepsilon)^2+\varepsilon^2\right)\ge0,$$ so condition $2$ is satisfied for $x\le1.$
On the other hand,
$$P_6(2\varepsilon) = \varepsilon,\quad \varphi(1+\varepsilon) = V(1+\varepsilon)$$ and $$\left(P_6(y) - (y+1-\varepsilon)\right)' = \dfrac1{16\varepsilon^5}\left(3y^5-15\varepsilon y^4 +20\varepsilon^2 y^3-16\varepsilon^5\right) = \dfrac1{16\varepsilon^5}\left(y-2\varepsilon\right)^3 (3y^2+3\varepsilon y+2\varepsilon^2)\le0\text{ if } y\in(\varepsilon, 2\varepsilon].$$ These mean that $$P_6(y)\ge y+1-\varepsilon, \text{ if } y\in(\varepsilon, 2\varepsilon],$$ so condition $2$ is satisfied for $x\in(1,1+\varepsilon],$
Proved.
You can try a connection polynomial with the following constraints:
$p(1-\epsilon)=p'(1-\epsilon)=p''(1-\epsilon)=0$,
$p(1+\epsilon)=\log(1+\epsilon),p'(1+\epsilon)=\dfrac1{1+\epsilon},p''(1+\epsilon)=-\dfrac1{(1+\epsilon)^2}$,
defined over $[1-\epsilon,1+\epsilon]$, and $V$ elsewhere.
As there are six constraints, you will need a quintic polynomial. The equations to determine the coefficients are linear.
There is no a priori guarantee that the polynomial will remain positive in the required range, but I wouldn't be surprised that this occurs naturally.