This question is #53 from the GRE Math Practice Book

Let f and g be functions of a real variable such that $g(x) = \int_{0}^{x} f(y)(y-x)dy$ for all x. If g is three times differentiable, what is the greatest integer n for which f must be n times continuously differentiable?

  • $\begingroup$ What have you thought about so far? $\endgroup$ Oct 12, 2015 at 15:53

1 Answer 1


Hint: It is easier to differentiate $$g(x)=\int_{0}^x f(y)y\,dy - x\int_{0}^x f(y)dy$$

What is $g'(x)$? What is $g''(x)$? $g'''(x)$?

  • $\begingroup$ Hmm, I keep wanting to say that g'(x) = 0 by using the Fundamental Theorem of Calculus and basically replacing y with x in each of the integrands and dropping the integrals. But then g would be infinitely differentiable. That first integral must require some sort of substitution..? Can you explain why or point me to some reading material for this? Thanks $\endgroup$
    – Ben
    Oct 12, 2015 at 16:10
  • $\begingroup$ @Ben $x\int_{0}^x f(y)\,dy$ is a product of the function $x$ and the integral. You can't just differentiate the integral. $\endgroup$ Oct 12, 2015 at 16:12
  • $\begingroup$ $$\left(\int_{a(x)}^{b(x)}f(t)dt\right)'=b'(x)f(b(x))-a'(x)f(a(x))$$ $\endgroup$
    – user5402
    Oct 12, 2015 at 16:12
  • $\begingroup$ Ohh, [facepalm]. Thank you. $\endgroup$
    – Ben
    Oct 12, 2015 at 16:13
  • $\begingroup$ @ThomasAndrews I was a typo. $\endgroup$
    – user5402
    Oct 12, 2015 at 16:15

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