What is the Range of a Differentiable Function with a Given Derivative at x=2?

[jd12345]

Homework Statement

A function from R-->R is differentiable and follows f( (x+y)/3 ) =( 2 + f(x) + f(y) ) / 3
Derivative of f(x) at x=2 is 2

Find the range of f ( |x| )

Homework Equations

The Attempt at a Solution

Well the questions asks me the range of f( |x| ). But i don't even know f(x). I did find the value of f(0) which is 2. And i don't know what to do afterwards. I have no idea why the value of derivative is given.

 

[SammyS]

Homework Statement

A function from R-->R is differentiable and follows f( (x+y)/3 ) =( 2 + f(x) + f(y) ) / 3
Derivative of f(x) at x=2 is 2

Find the range of f ( |x| )

Homework Equations

The Attempt at a Solution

Well the questions asks me the range of f( |x| ). But i don't even know f(x). I did find the value of f(0) which is 2. And i don't know what to do afterwards. I have no idea why the value of derivative is given.

How is f(-x) related to f(x) ?

Spoiler

Hint: Let y = -x .

Use that result to see how f '(-x) is related to f '(x) .

 

[SammyS]

Two more substitutions that might be helpful:
y = x (which you may have used to fin f(0) )

and

y = 2x .

 

[jd12345]

Ok (by using spoiler) f(0) = 2 + f(x) + f(-x) / 3
=> 6 = 2 + f(x) +f(-x)
=> 4 = f(x) +f(-x)
Thus 0 = f '(x) - f '(-x)
f '(x) = f '(-x)
f '(x) is even so f(x) is odd.

But if f(x) is odd then f(x) + f (-x) = 0 but i found out earlier that it is equal to 4. A bit confused

 

[SammyS]

Ok (by using spoiler) f(0) = 2 + f(x) + f(-x) / 3
=> 6 = 2 + f(x) +f(-x)
=> 4 = f(x) +f(-x)
Thus 0 = f '(x) - f '(-x)
f '(x) = f '(-x)
f '(x) is even so f(x) is odd. This is not true.

But if f(x) is odd then f(x) + f (-x) = 0 but i found out earlier that it is equal to 4. A bit confused

If f(x) is odd, then it is true that f '(x) is even, but the converse is not necessarily true.

If f '(x) is odd, then it does follow that f(x) is even.

In your case, 4 = f(x) +f(-x), so that f(-x) = -f(x) + 4 . Therefore, f(x) is not odd.

Consider the function g(x) defined as g(x) = f(x) - 2.

g(x) is odd.

 

[jd12345]

Oh ok - but still i cannot solve the initial problem. This is what i have done till now:-
f(0) = 2
f(x) = f(-x) + 4 and f '(x) = f ' (-x)
Are they right? and what else do i need to do?

 

[SammyS]

Two more substitutions that might be helpful:
y = x (which you may have used to find f(0) )

and

y = 2x .

See what derivative relationships result from the above.

Also, you might consider investigating the behavior of the function g(x) .