I could not get started with this question. Perhaps it is very simple.
27. Let f be a continuous function such that f(x) = f(1  x) for all real numbers x. If f is differentiable everywhere, then f'(0) =?
a. f(0)
b. f(1)
c. f(0)
d. f'(1)
e. f(1) this is the answer
thanks
GR9768 problem 27
Ok! just use the chain rule in doing the derivative. i.e. if y=f(g(x)) ==> y'=g'(x)*f'(g(x)). Now see how this rule applies to the question.
let g(x)=1x, now [f(1x)]'=f'(x). Finally set x=0 and obtain. f'(0) [comes from the LHS of equation] = f'(1) [which comes from the RHS of the equation].
That's all. :d
let g(x)=1x, now [f(1x)]'=f'(x). Finally set x=0 and obtain. f'(0) [comes from the LHS of equation] = f'(1) [which comes from the RHS of the equation].
That's all. :d

 Posts: 19
 Joined: Fri Mar 04, 2016 7:37 pm
Re: GR9768 problem 27
Another way is to see that f(x) is an even function shifted by 1/2 to the right. As 0 and 1 are equidistant to 1/2, their derivatives are opposites, one is negative while the other is positive, or vice versa.
Sketch y = 1/2  x to illustrate.
Sketch y = 1/2  x to illustrate.