If we can find a function f(x) such that f(2)=20, f(6)=10 and f(c)=25 where c is in [2,6], then we disprove the statement.
An inverted parabola would fit the criteria, for example: f(x)=-2.5x^2+17.5x-5. In this function f(3)=f(4)=25. So the statement is false.
The IVT doesn't cover all possible transitions in the range. Similar to the statement that all squares are polygons but not all polygons are squares. The IVT simply says that there will be an intermediate value in the domain that produces a value in the range between f(2) and f(6). f(5.5)=15.625, which is indeed between 20 and 10. It doesn't exclude values higher than both the end points.
I worked out the solution by assuming that f(x) could be expressed as ax^2+bx+c, where a, b, c are constants. Then I assumed that an intermediate value for x would produce f(x)=25. I chose an abitrary value of 4 (midrange). This gave me a system of equations from which a, b, c could be calculated.