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GR9768 problem 27

 
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fullofquestions



Joined: 07 Oct 2007
Posts: 18
Posted: Tue Oct 30, 2007 2:20 pm    Post subject: GR9768 problem 27
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
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mamal



Joined: 30 Oct 2007
Posts: 11
Posted: Thu Nov 01, 2007 7:02 am    Post subject:
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)=1-x, now [f(1-x)]'=-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
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