Thread: How to solve: If f(x) is even fcn, find range of b. f(x)=[3.5+bsinx] where [] is gif

1. Originally Posted by Abhishekdas
becoz [3.5+bsinx] is not equal to f(x) = 3.5 + b * sin(x)
You answered that f(x) is not even because $\lfloor{f(x)}\rfloor \ne f(x)$.

It's true that f(x) is not even, but I don't think your reason why is valid.

f(x) is not even because sin(x) is not even.

Consider the following function g.

g(x) = 3.5 + b*cos(x)

This is an even function because cos(x) is an even function.

If we let b = 1, for an example, here's the graph of $g(x)$ and $\lfloor{g(x)}\rfloor$.

$\lfloor{g(x)}\rfloor \ne g(x)$, yet g(x) is even.

2. I don't get it

Originally Posted by Subhotosh Khan
Then do you not see that the only way f(x) can become an even function is when b = 0
I don't get it. Isn't f(x)= [3.5+bsinx]. Kindly elaborate.

3. Originally Posted by Abhishekdas
I don't get it. Isn't f(x)= [3.5+bsinx]. Kindly elaborate.
Yes, that is the definition of f(x). That can't be questioned as it was given!

YOU said that a function f(x) is defined to be an even function if f(x) = f(-x), yet you claimed, as mbot pointed out, that f(x) is not even because
[f(x)] is not equal to f(x). Where did this definition come from????

You want to verify whether or not f(x)=f(-x), that is whether or not [3.5+bsin(x)] = [3.5+bsin(-x)] . When is this true? Remember that your answer should depend on b

4. Originally Posted by Subhotosh Khan
Then do you not see that the only way f(x) can become an even function is when b = 0
Hmmm, how about |b|< 1/2 ???????

5. Originally Posted by Jomo
Hmmm, how about |b|< 1/2 ???????
You are correct for OP.

However, I was trying to "nudge" OP in that direction by assuming a simpler function (removing GIF operation).