Let $f : \mathbb{R} \to \mathbb{R}$ be a function with continuous derivative such that $f(\sqrt{2}) = 2$ and $$f(x) = \lim_{t \to 0}\dfrac{1}{2t}\int_{x-t}^{x+t}sf'(s)\,ds \ \text{for all} \ x \in \mathbb{R}.$$ Then $f(3)$ equals

$(a) \ \ \sqrt{3} \hspace{1.25 in} (b) \ \ 3\sqrt{2} \hspace{1.25 in} (c) \ \ 3\sqrt{3} \hspace{1.25 in} (d) \ \ 9$

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Can someone help me solve this problem on real function . Here is my try I differentiated both sides with respect to $x$. And I got derivative of $x$ to be $0$. I concluded $f(x)$ is a constant function. So answer should be $2$. But $2$ is not an option . How to solve this ??


1 Answer 1


We have: \begin{align} f(x)&=\lim_{t \rightarrow 0}\frac{1}{2t}\int_{x-t}^{x+t}sf'(s)\,ds\\ &=\lim_{t \rightarrow 0}\frac{1}{2t}sf(s)\,\Big|_{x-t}^{x+t}-\lim_{t \rightarrow 0}\frac{1}{2t} \int_{x-t}^{x+t}f(s)\,ds\\ &=\lim_{t \rightarrow 0}\frac{1}{2t}sf(s)\,\Big|_{x-t}^{x+t}-f(x) \end{align} or \begin{align} 2f(x)&=\lim_{t \rightarrow 0}\frac{1}{2t}s f(s)\,\Big|_{x-t}^{x+t}\\ &=\lim_{t \rightarrow 0}\frac{1}{2t}\cdot \left[ (x+t)f(x+t) - (x-t)f(x-t) \right]\\ &=\lim_{t \rightarrow 0}\frac{1}{2t}\cdot \left[ x(f(x+t) - f(x-t)) +t(f(x+t) + f(x-t)) \right]\\ &=x \lim_{t \rightarrow 0}\frac{f(x+t) - f(x-t)}{2t} + \lim_{t \rightarrow 0} \frac{f(x+t) + f(x-t)}{2}\\ &=xf'(x) + f(x) \end{align} Or $$f(x)=xf'(x)$$ which is $$f(x)=c x$$ or, from $f(\sqrt{2})=2$, $$f(x)=\sqrt{2} x$$ the answer is (b)

  • 2
    $\begingroup$ You may more easily obtain $f(x)=x f'(x)$ by noting that $ \lim_{t\to 0}(1/t)\int_x^{x+t}sf'(s)ds=$ $\lim_{t\to 0}(1/t)\int_{x-t}^xsf'(s)ds$ $=xf'(x).$ $\endgroup$ Dec 24, 2016 at 0:46
  • $\begingroup$ Yep, that's a nice shortcut indeed, thank you @user254665! $\endgroup$
    – rtybase
    Dec 24, 2016 at 0:53
  • $\begingroup$ What is the first step you have done $\endgroup$ Dec 24, 2016 at 1:54
  • $\begingroup$ Is it a formula? $\endgroup$ Dec 24, 2016 at 1:55
  • $\begingroup$ Thanks I get it you have used by parts $\endgroup$ Dec 24, 2016 at 1:58

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