Differentiable function f for x > 0

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dws

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Mar 1, 2009
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Hi, I don't know how to handle this problem. Help is appreciated.

Let f be a differentiable function defined for all x > 0 such that

i. f(1) = 0

ii. f '(1) = 1, and

iii. (d/dx)[f(2x)] = f '(x), for all x > 0. (left hand side of the equation is d over dx to the left of [f(2x)] )

a) Find f '(2).

f '(2x) = f '(x) for x > 0. f '(1) = 0, so f '(2) = 1.

b) Suppose f ' is differentiable. Prove that there is a number c, 2 < c < 4, such that

f ''(c) = -1/8

I know I need to use the MVT, but I don't know more than that.

c) Prove that f(2x) = f(2) + f(x) for all x > 0

No clue here. Integrate, differentiate?
 
I assumed that to be true, but where does f(2) come in to replace c? I found f(x) = x - 1 by assuming constant slope of 1 and using f(1) = 0. That satisfies the conditions, but f(2x) isn't f(x) + f(2) in that case. Where to go from there?
 
dws said:
Hi, I don't know how to handle this problem. Help is appreciated.

Let f be a differentiable function defined for all x > 0 such that

i. f(1) = 0

ii. f '(1) = 1, and

iii. (d/dx)[f(2x)] = f '(x), for all x > 0. (left hand side of the equation is d over dx to the left of [f(2x)] )

a) Find f '(2).

f '(2x) = f '(x) for x > 0. f '(1) = 0, so f '(2) = 1. << I believe you have some typos in your problem statement
b) Suppose f ' is differentiable. Prove that there is a number c, 2 < c < 4, such that

f ''(c) = -1/8

I know I need to use the MVT, but I don't know more than that.

c) Prove that f(2x) = f(2) + f(x) for all x > 0

No clue here. Integrate, differentiate?
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