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