Is the sum of two functions, one of which is non differentiable, necessarily non-differentiable? Suppose $f(x) = g(x) + h(x)$ where $g(x)$ is known to be non-differentiable. When $h(x) \ne -g(x)$ is some other function (differentiable or not), can $f(x)$ ever be differentiable? We can assume $f, g, h$ are real valued functions.
edit: I should have specified that $h(x)$ is not some function that cancels out $g(x)$ . Bill Dubuque's answer is intriguing.
 A: Yes, of course. For example, if $h(x)=1-g(x)$, then $f(x)=1$ is differentiable.
A: Easy counterexamples have been mentioned. What is true, however, is the fact that sum of a differentiable function and a nondifferentiable function is nondifferentiable. This is true because differentiable functions are closed under subtraction, i.e. they comprise a subgroup of all functions. Hence the claimed property is simply a special case of the following complementary form of the subgroup property from my prior post.
THEOREM $\  $ A nonempty subset $\rm\:S\:$ of abelian group $\rm\:G\:$ 
comprises a subgroup $\rm\iff\ S\ + \ \bar S\ =\ \bar S\ $ where  $\rm\: \bar S\:$ is the complement of $\rm\:S\:$ in $\rm\:G$
Instances of this are ubiquitous in concrete number systems, e.g. 

                                 transcendental 
     algebraic * nonalgebraic  =  nonalgebraic  if  nonzero 
      rational * irrrational   =   irrational   if  nonzero 
          real *   nonreal     =    nonreal     if  nonzero 

         even  +     odd       =      odd          additive example
       integer + noninteger    =   noninteger


A: And to answer the other side of the question (is the function of a continuous function and a discontinuous function always discontinuous?), suppose that $f(x) = g(x)+h(x)$ is differentiable as before, with $g(x)$ non-differentiable and $h(x)$ differentiable.  Then $g(x) = f(x) - h(x)$ is the difference of differentiable functions, and it's a pretty quick (and standard) exercise in deltas and epsilons to show that this must be differentiable itself.
A: Let $h(x)=-g(x)+k(x)$ where $k\neq0$ is any differentiable function other than the constant 0 function. Then $f(x)=g(x)-g(x)+k(x)=k(x)$ is differentiable.
