Solving Logarithmic Series Homework: Find x Values

[ritwik06]

Homework Statement

$$x=\frac{1+\frac{f(x+1)}{1!}+\frac{f^{2}(x+1)}{2!}+\frac{f^{3}(x+1)}{3!}+...}{1+\frac{f(x)}{1!}+\frac{f^{2}(x)}{2!}+\frac{f^{3}(x)}{3!}+...}$$
f(x) is a twice differentaible equation.
1. Find the possible values of x when;
f(x+0.5)<f(x-0.5)
2. Find the possible values of x when;
f(|x|+e-1)<f(|x|+e-2)+1

The Attempt at a Solution

On simplifying i get:
ln x= f(x+1)- f(x)
The only other data given is that f(x) is twice differentiable, which just means that when the function is expressed it will have two distinct constants, right? How shall I proceed?

 

[HallsofIvy]

Why would "twice differentiable" say anything about constants?

 

[ritwik06]

Why would "twice differentiable" say anything about constants?

The order of differential equation is equal to number of distinct arbitrary constants. Right?
If not, tell me how can I do these?

I have shown everything I knew! Please help!

 

[HallsofIvy]

There is no differential equation here!

x+ 1/2= x- 1/2+ 1 so f(x+ 1/2)= f((x-1/2)+1). Saying that f(x+1/2)< f(x- 1/2) means that f((x-1/2)+ 1)< f(x- 1/2) or f(y+ 1)< f(y) with y= x- 1/2. So what is ln(y) and what does that tell you about x?