Well, "to start the ball rolling", let us assume the desired function is the second-degree polynomial \begin{equation} f(x) =a_0 +a_1 x +a_2 x^2. \end{equation} Then, we can achieve the three target exact results $f(\frac{1+t}{2})=\frac{25}{341}$, $f(\sqrt{t}) =1 -\frac{256}{27 \pi^2}$ and $f(\frac{2 t}{1+t}) =0$, by taking \begin{equation} a_0=-\frac{2 \sqrt{t} \left(27 \pi ^2 \left(682 t^{3/2}+341 t^2+582 t+682 \sqrt{t}+341\right)-87296 \left(\sqrt{t}+1\right)^2 (t+1)\right)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2}, \end{equation} \begin{equation} a_1=\frac{27 \pi ^2 (t (341 t+1946)+341)-87296 (t (t+6)+1)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \sqrt{t}}, \end{equation} and \begin{equation} a_2=-\frac{2 (t+1) \left(27 \pi ^2 \left(341 t+632 \sqrt{t}+341\right)-87296 \left(\sqrt{t}+1\right)^2\right)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2 \sqrt{t}}, \end{equation} giving us that the second-degree polynomial $f(x)$ is obtainable by dividing \begin{equation} \left(27 \pi ^2 \left(682 t^{3/2}+341 t^2+t (582-682 x)+\sqrt{t} (682-1264 x)-682 x+341\right)-87296 \left(\sqrt{t}+1\right)^2 (t-2 x+1)\right) (t (x-2)+x) \end{equation} by \begin{equation} 9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2 \sqrt{t}. \end{equation}

Our four other target values are only numerical and perhaps accurate to only 3-4 decimal places. Rationalizing one of them, 0.0346801, we added $f(\frac{1}{4} (\sqrt{t}+1)^2=\frac{347}{10000}$ to our set of equations, and obtained a further (larger) solution.

In this proof-of-principle exercise, one could take $f(x)$ to be other than polynomial in nature--rational functions,...

However, perhaps in my original conception of the problem I was thinking that $t$ would not be present in the expression of the desired function $f(x)$, as it certainly is in the coefficients of the second-degree polynomial given above.

Well, "to start the ball rolling", let us assume the desired function is the second-degree polynomial \begin{equation} f(x) =a_0 +a_1 x +a_2 x^2. \end{equation} Then, we can achieve the three target exact results $f(\frac{1+t}{2})=\frac{25}{341}$, $f(\sqrt{t}) =1 -\frac{256}{27 \pi^2}$ and $f(\frac{2 t}{1+t}) =0$, by taking \begin{equation} a_0=-\frac{2 \sqrt{t} \left(27 \pi ^2 \left(682 t^{3/2}+341 t^2+582 t+682 \sqrt{t}+341\right)-87296 \left(\sqrt{t}+1\right)^2 (t+1)\right)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2}, \end{equation} \begin{equation} a_1=\frac{27 \pi ^2 (t (341 t+1946)+341)-87296 (t (t+6)+1)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \sqrt{t}}, \end{equation} and \begin{equation} a_2=-\frac{2 (t+1) \left(27 \pi ^2 \left(341 t+632 \sqrt{t}+341\right)-87296 \left(\sqrt{t}+1\right)^2\right)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2 \sqrt{t}}, \end{equation} giving us that the second-degree polynomial $f(x)$ is obtainable by dividing \begin{equation} \left(27 \pi ^2 \left(682 t^{3/2}+341 t^2+t (582-682 x)+\sqrt{t} (682-1264 x)-682 x+341\right)-87296 \left(\sqrt{t}+1\right)^2 (t-2 x+1)\right) (t (x-2)+x) \end{equation} by \begin{equation} 9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2 \sqrt{t}. \end{equation}

Our four other target values are only numerical and perhaps accurate to only 3-4 decimal places. Rationalizing one of them, 0.0346801, we added $f(\frac{t-1}{\log{t}})=\frac{347}{10000}$ to our set of equations, and obtained a further (larger) solution.

In this proof-of-principle exercise, one could take $f(x)$ to be other than polynomial in nature--rational functions,...

However, perhaps in my original conception of the problem I was thinking that $t$ would not be present in the expression of the desired function $f(x)$, as it certainly is in the coefficients of the second-degree polynomial given above.

Well, "to start the ball rolling", let us assume the desired function is the second-degree polynomial \begin{equation} f(x) =a_0 +a_1 x +a_2 x^2. \end{equation} Then, we can achieve the three target exact results $f(\frac{1+t}{2})=\frac{25}{341}$, $f(\sqrt{t}) =1 -\frac{256}{27 \pi^2}$ and $f(\frac{2 t}{1+t}) =0$, by taking \begin{equation} a_0=-\frac{2 \sqrt{t} \left(27 \pi ^2 \left(682 t^{3/2}+341 t^2+582 t+682 \sqrt{t}+341\right)-87296 \left(\sqrt{t}+1\right)^2 (t+1)\right)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2}, \end{equation} \begin{equation} a_1=\frac{27 \pi ^2 (t (341 t+1946)+341)-87296 (t (t+6)+1)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \sqrt{t}}, \end{equation} and \begin{equation} a_2=-\frac{2 (t+1) \left(27 \pi ^2 \left(341 t+632 \sqrt{t}+341\right)-87296 \left(\sqrt{t}+1\right)^2\right)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2 \sqrt{t}}, \end{equation} giving us that the second-degree polynomial $f(x)$ is obtainable by dividing \begin{equation} \left(27 \pi ^2 \left(682 t^{3/2}+341 t^2+t (582-682 x)+\sqrt{t} (682-1264 x)-682 x+341\right)-87296 \left(\sqrt{t}+1\right)^2 (t-2 x+1)\right) (t (x-2)+x) \end{equation} by \begin{equation} 9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2 \sqrt{t}. \end{equation}

Our four other target values are only numerical and perhaps accurate to only 3-4 decimal places. Limited attempts of ours to increase the degree of $f(x)$ and fit exact rational approximations to these values, along with the three original exact values, have not been successful.

In this proof-of-principle exercise, one could take $f(x)$ to be other than polynomial in nature--rational functions,...

Well, "to start the ball rolling", let us assume the desired function is the second-degree polynomial \begin{equation} f(x) =a_0 +a_1 x +a_2 x^2. \end{equation} Then, we can achieve the three target exact results $f(\frac{1+t}{2})=\frac{25}{341}$, $f(\sqrt{t}) =1 -\frac{256}{27 \pi^2}$ and $f(\frac{2 t}{1+t}) =0$, by taking \begin{equation} a_0=-\frac{2 \sqrt{t} \left(27 \pi ^2 \left(682 t^{3/2}+341 t^2+582 t+682 \sqrt{t}+341\right)-87296 \left(\sqrt{t}+1\right)^2 (t+1)\right)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2}, \end{equation} \begin{equation} a_1=\frac{27 \pi ^2 (t (341 t+1946)+341)-87296 (t (t+6)+1)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \sqrt{t}}, \end{equation} and \begin{equation} a_2=-\frac{2 (t+1) \left(27 \pi ^2 \left(341 t+632 \sqrt{t}+341\right)-87296 \left(\sqrt{t}+1\right)^2\right)}{9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2 \sqrt{t}}, \end{equation} giving us that the second-degree polynomial $f(x)$ is obtainable by dividing \begin{equation} \left(27 \pi ^2 \left(682 t^{3/2}+341 t^2+t (582-682 x)+\sqrt{t} (682-1264 x)-682 x+341\right)-87296 \left(\sqrt{t}+1\right)^2 (t-2 x+1)\right) (t (x-2)+x) \end{equation} by \begin{equation} 9207 \pi ^2 \left(\sqrt{t}-1\right)^4 \left(\sqrt{t}+1\right)^2 \sqrt{t}. \end{equation}

Our four other target values are only numerical and perhaps accurate to only 3-4 decimal places. Rationalizing one of them, 0.0346801, we added $f(\frac{1}{4} (\sqrt{t}+1)^2=\frac{347}{10000}$ to our set of equations, and obtained a further (larger) solution.

In this proof-of-principle exercise, one could take $f(x)$ to be other than polynomial in nature--rational functions,...

However, perhaps in my original conception of the problem I was thinking that $t$ would not be present in the expression of the desired function $f(x)$, as it certainly is in the coefficients of the second-degree polynomial given above.