Since f(x) is continuous, so is h(x).
Also, since f (3) = 4 and f (5) = 2, we have
h (3) = f (3) + 2•3 = 4 + 6 = 10
h (5) = f (5) + 2•5 = 2 + 10 = 12
It follows from the intermediate value theorem that there is some t between 3 and 5 such that h(t) is between 10 and 12, and namely there is some t for which h(t) = 11.
