$f$ is convex and $f(10)$, $f(20)$ given. Find the smallest value of $f(7)$.

You must have $$f(10) =fleft(frac{10}{13} times 7 + frac{3}{13} times 20right) leq frac{10}{13} times f(7) + frac{3}{13} times f(20)$$

So $$-4 leq frac{10}{13} times f(7) + frac{3}{13} times 30$$

i.e. $$f(7) geq -frac{71}{5} $$

Now you can check that it is the lower bound. Indeed, the function $$f : x mapsto frac{17}{5}x - 38$$

is convex, satisfies $f(10)=-4$ and $f(20)=30$, and $f(7)=-frac{71}{5}$.

Hint: $10=a(7)+(1-a)(20)$ where $a=frac {10} {13}$. Hence $-4=f(10) leq af(7)+(1-a)(30)$. This gives lower bound for $f(7)$ and this value is attained when $f$ is a linear function. [ If you draw the straight line passing through the points $(10,-4), (20,30)$ then you get the graph of a convex function whose value at $7$ is $-frac {71} 5$ the lower bound you get from above inequality].