Multiply your function with an infinitely differentiable function phi(epsilon x) where phi(x) is 0 for x less or equal to 0 and equal to 1 for x greater than 1 as is defined in the link given below. Then, the product is infinitely differentiable and differs from your function only in the region 0< x < epsilon than is tiny for small epsilon.

http://www.math.toronto.edu/mat1300/partitions.10.pdf

I think you can use smooth transition functions. Check Lstar models (logistic Smooth trnsition autorregrive models. you can also check my Ph. D. thesis "Asymmetries in business cycles". I use that kind of function to make differentiable a kink utility funcion, Specifically check chapter 2.

 @Herbert H H Homeier

Sir could you explain a bit more. I am cofused  by phi(epsilonx)

Sorry for not really taking full respect to the original question. There you had a nondifferentiality at x=9. In my answer I implied a shift of the function such that the nondifferentiality is at x=0. Then, "epsilon times x" is what is meant to be the argument of phi for the shifted situation. This is necessary to scale from [0,1] to [0,epsilon] for the interval where the product of functions differs from the shifted function F(x+9) for my position x=0 of the nondifferentiality. I should have used a different variable X, say, in my answer.... For the unshifted case you can multiply F(x) by phi(epsilon (x - 9) ). I recommend to plot the functions F and the product in some interval of length 4 epsilon with center at 9 for some small epsilon in order to see what is going on. Good luck

@Herbert H H Homeier 

Thanks a lot Sir for your valuable advice.

Your idea works fine in my case. The approximation phi(epsilon (x - 7.75) ) works well for a value of epsilon around epsilon=0.2. I am attaching the comparative graph of the original and new functions for different values of epsilon(=0.9,0.5,0.2). It seems that for epsilon=0.9 the graph is approximated by some straight lines approximately around x=(7.75,8.9). But for epsilon=2 the graph is smooth.

Kindly comment if the approximation is appropriate or not. The tolerance of error in y(x) is

Since you already got a method that worked, I shouldn't comment.  But I suggest a general method. You may consider an epsilon neighborhood of the point of discontinuity,  where epsilon is presumably small and then make epsilon tend to zero finally.