Differentiable function

Question

Let f be a differentiable function defined for all x>0 such that (i) f(1) = 0, (ii) f'(1) = 1 and (iii) d[f(2x)]/dx = f'(x), for all x>0.

(a) find f'(2)

(b) suppose f' is differentiable, prove that there is a number c, 2<c<4, such that f"(c) = -1/8.

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

Answer 1

Given: f(1) = 0 and f'(1) = 1 and d[f(2x)] / dx = f'(x) for all x > 0

a. Using the Chain Rule, d[f(2x)] / dx = f'(2x)(2) = f'(x) for all x > 0

So, when x = 1, we have 2f'(2) = f'(1) = 1.

Therefore, f'(2) = 1/2.

b. By the Mean Value Theorem, there is a number c in the interval (2,4) such that

f"(c) = (f'(c))' = [f'(4) - f'(2)] / (4 - 2)

From part a, f'(2) = 1/2

f'(2x)(2) = f'(x), So, when x = 2, f'(4)(2) = f'(2).

Therefore, f'(4) = 1/4

So, f"(c) = [1/4 - 1/2] / 2 = (-1/4) / 2 = -1/8

c. From the above, we know that f'(2x)(2) = f'(x)

So, 2∫f'(2x)dx = ∫f'(x)dx

Let u = 2x. Then du = 2dx. So, dx = (1/2)du.

Therefore, 2∫f'(u)(1/2)du = ∫f'(x)dx

f(u) = f(x) + C

f(2x) = f(x) + C

So, f(2) = f(1) + C = 0 + C = C

Therefore, f(2x) = f(x) + f(2)