Solving Logarithmic Series Homework: Find x Values

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Homework Help Overview

The problem involves finding values of x based on inequalities involving a function f(x), which is described as twice differentiable. The inequalities compare the function's values at different points, and the original poster expresses confusion regarding the implications of the function's differentiability and the nature of the problem.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to simplify the problem and relates the differentiability of f(x) to the presence of constants in its expression. Some participants question this reasoning, specifically the connection between differentiability and constants. Others explore the implications of the inequalities presented.

Discussion Status

The discussion is ongoing, with participants raising questions about the nature of the problem and the assumptions made regarding the function f(x). There is no explicit consensus, but participants are engaging with the concepts and attempting to clarify the original poster's understanding.

Contextual Notes

There is a lack of clarity regarding the relationship between the function's differentiability and the constants involved. Additionally, the original poster's interpretation of the problem as involving a differential equation is being challenged.

ritwik06
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Homework Statement



[tex]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!}+...}[/tex]
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?
 
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Why would "twice differentiable" say anything about constants?
 
HallsofIvy said:
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!
 
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?
 

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