From the problems plus in Stewart Calculus 6e, it asks if $f$ is a differentiable function such that $f(x)$ is never $0$ and for any $x$, $\int_0^xf(t)dt=[f(x)]^2$, then what is $f(x)$?

I figured since it's differentiable I could take the derivative of both sides to get:
$$f(x)=2f(x)f'(x)$$ Since $f(x)$ is never $0$, then $f'(x)=1/2$. But that means that $f(x)=x/2+c$, which will equal $0$ for $x=-2c$. I can't just take $0$ out of the function, since it has to be differentiable everywhere. So how do I solve this problem?

  • $\begingroup$ You already have. If $f(x)\not =0$ then it follows that $f(x) = \frac{x}{2}$, but then $f(0) = 0$ so unless you have misread the problem you have showed that there are no solutions that satisfy $f(x)\not =0$ for all $x$. $\endgroup$ – Winther Aug 10 '15 at 23:35
  • $\begingroup$ Also, at $x = 0$ the LHS is $0$, while the RHS is $(f(0))^2$. Perhaps the problem is asking for $f$ that is not identically zero, i.e. it is not the case that $f(x) = 0$ for all $x$. $\endgroup$ – Joey Zou Aug 10 '15 at 23:40
  • $\begingroup$ Thanks for the reassurance, that was what I was thinking. But still, the problem reads exactly: "If $f$ is a differentiable function such that $f(x)$ is never 0 and $\int_0^xf(t)dt=[f(x)]^2$ for all $x$, find $f$." Ah well, it must just be a mistake. $\endgroup$ – Sam Jaques Aug 11 '15 at 3:49

Observe that $f(x) > 0$ for $x > 0$. Thus $D_f = \{x: x > 0\}$. On $D_f$, we have: $f(x)(1-2f'(x)) = 0 \to f'(x) = \dfrac{1}{2} \to f(x) = \dfrac{x}{2} + C \to C = \displaystyle \lim_{x\to 0} f(x) = 0 \to f(x) = \dfrac{x}{2}$


I believe that you have misread the problem. I think when it says $f(x) \neq 0$ means that $f(x)$ is not the zero function (i.e. it is not always zero). Otherwise you seem to have solved the problem.

EDIT: I just read your comment. I think your book either has a misprint or the correct answer is "there is no solution". When $x=0$, we get $0=f(0)^2$ so $f(0)=0$, so no matter what $f(x)$ must equal $0$ at some point.

  • $\begingroup$ that is a good point. huge difference between $f(x)\neq 0$ and $f(x)\neq 0\ \ \forall x$ $\endgroup$ – Elliot G Aug 11 '15 at 4:33

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