Answer
$f(x) = cx+d~~$ for some constant $d$
Work Step by Step
Let $g(x) =cx$
Then $g'(x) = c$
$f'(x) = g'(x) = c~~~~$ for all $x$
According to Corollary 7, $f(x) = g(x)+d$ for some constant $d$
Thus, $f(x) = g(x)+d = cx+d~~$ for some constant $d$