2
$\begingroup$

I was taking a test and two true/false type questions were asked.

In one of them, I had to say if there is a function $f(x)$ such that $f(x) = f'(x)$. Of course, $e^x$ is such a function and almost everyone who has taken a calculus course knows this fact well.

In the other question, I had to determine if $f(x) = -f'(x)$ was possible.

I was completely stumped at this one. I had never before encountered a function with such property nor did I know how to approach this problem analytically as I am just a high school student.

My question is: is there an analytical way to determine if such a function exists? By analytical, I mean no guessing allowed and just giving an example won't be enough.

Is this possible? If not, can you give an example of a function with the above property?

$\endgroup$
  • $\begingroup$ Also, $f(x)=0$. $\endgroup$ – Micapps Feb 4 '19 at 9:40
  • $\begingroup$ If you know the formula $\frac d {dx} [f(-x)]=-[\frac d {dx} f](-x)$ then there is not much to guess. $\endgroup$ – Kavi Rama Murthy Feb 4 '19 at 9:41
3
$\begingroup$

If $f'(x)=f(x)$ and if $g(x)=f(-x)$, then $g'(x)=-f'(-x)=-f(-x)=-g(x)$. Can you take it from here?

$\endgroup$
4
$\begingroup$

My question is: is there an analytical way to determine if such a function exists?

There's a theorem for that. Specifically, the existence-uniqueness theorem for differential equations.

Wikipedia link

$\endgroup$
0
$\begingroup$

Linear differential equations $ay"+by'+cy$ are solved by considering the roots of the polynomial $ax^2+bx+c$, here you have $x+1=0$.

$\endgroup$
0
$\begingroup$

As you're in high school, you have probably not covered the topic of differential equations but you can use one to find the analytical solution to your question. $dy/dx + y = 0$ and you will find that the solution is $y=ce^{-x}$ (where c is any constant): https://www.wolframalpha.com/input/?i=dy%2Fdx+%2B+y+%3D+0

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.