# Fundamental theorem of calculus question from GRE

This question is motivated from GRE math subject test.

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

$$\text{(a) } 1\quad\text{(b) } 2\quad\text{(c) } 3\quad\text{(d) } 4\quad\text{(e) } 5$$

What I thought is to try differentiating the function. By the fundamental theorem of calculus I thought that $$\mathrm{d}g/\mathrm{d}x$$ should be $$f(x)(x-x)=0$$ which I guess is wrong so I figured that FTC must not work for multi-variable functions. So I have no idea what to do. Does anyone have any ideas?

• Just saying that you're missing an assumption, that $f$ is continuous. $g$ can be smooth without $f$ being continuous.
– user223391
Oct 9, 2016 at 14:32
• Is the answer 2 Oct 9, 2016 at 14:42

First of all if $f(x)=\chi_{\{0\}}(x)$ then $g\equiv 0$ which is smooth so we must assume $f$ is continuous.

Note: $g(x)=\int_0^x f(y)(y)\ dy-x\int_0^x f(y)\ dy$

Then differentiate by FTC:

$g'(x)=f(x)x-xf(x) -\int_0^x f(y)\ dy=-\int_0^xf(y)\ dy$

You can differentiate again by FTC:

$g''(x)=-f(x)$

Then to differentiate again we need $f$ to be $C^1$ so the answer is $1$.