GR9768 problem 27

Forum for the GRE subject test in mathematics.
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GR9768 problem 27

Postby fullofquestions » Tue Oct 30, 2007 10:20 am

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


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Postby mamal » Thu Nov 01, 2007 3:02 am

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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Joined: Fri Mar 04, 2016 7:37 pm

Re: GR9768 problem 27

Postby danieloliveira56 » Wed Apr 13, 2016 8:32 pm

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.

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