# Is there a function such that $f(x)=f'(x)$?

I am asked to prove and explain that this statement is False? $$\exists f(x): f(x)=f'(x)$$

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What about exponential function? –  Kaster Jul 16 at 18:54
It's not always true, but sometimes it is. –  user67258 Jul 16 at 18:59
@SaraSharp -1 "that's all I have as a question" sounds like you just copied a homework problem word for word without even understanding what it was asking, let alone attempting it. –  Ataraxia Jul 16 at 19:00
The question in the title is different to that in the body of the question, and the latter is more or less meaningless (it is false for all $f$'s, for some $f$'s...?) –  Mariano Suárez-Alvarez Jul 16 at 19:10
Cool down folks. –  amWhy Jul 16 at 19:10

But why would you want to prove it is false? It happens to be very true. To prove the existence of a function such that $f(x) = f'(x)$, all you need to do in your "proof" is to provide/name one such function $f(x)$ for which $f(x) = f'(x)$.

Consider $f(x) = Ce^x$ ${}{}{}{}{}{}{}$

$$f(x) = Ce^x\quad \text{and}\quad f'(x) = Ce^x$$

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I have to prove if the statement is true or false. –  Sara Sharp Jul 16 at 19:00
Sara: All you need to do to prove the existence of a function (hence prove that the statement is true) is to provide even just one example of such a function $f(x) = f'(x)$. That's it! –  amWhy Jul 16 at 19:01
So, for example, since $C$ can be ANY real number, including $0$, your proof will consist of providing some example of a function, say, let $f(x) = 2e^x$. Then since $f'(x) = 2e^x$, we have that $f(x) = f'(x) = 2e^x$. Letting C = 0 is even simpler yet! –  amWhy Jul 16 at 19:05
Thanks amWhy your explanations are extremely helpfull. It was a confusing statement. Because, answer was telling me has to be false. –  Sara Sharp Jul 16 at 19:08
It's a little lame that you did not respond to Git Gud comment (yes, you edit your answer in the meantime (I saw that), but....). –  Cortizol Jul 16 at 19:32

The statement is True. All we need to do to show this is to find one function that satisfies $f(x)=f'(x)$. The simplest such function is $f(x)=0$.

More generally, any function of the form $f(x)=Ce^x$, where $C$ is a real constant, satisfies $f(x)=f'(x)$. If fact, this is the only function for which this equality holds.

To see this, let $y=f(x)$ so that $\frac{dy}{dx} = f'(x)$. We have that,

\begin{align} f(x) &= f'(x) \\ y &= \frac{dy}{dx} \\ \frac{dy}{y} &= dx \\ \int \frac{dy}{y} &= \int dx \\ \ln{y} &= x + c \\ e^{\ln{y}} &= e^{x+c} \\ y=e^{x+c} &= e^c e^x = Ce^x \end{align}

So $f(x)=y=Ce^x$.

It should be noted that $f(x) = 0$ is just the special case where $C = 0$ so that $$f(x) = 0 \cdot e^x = 0.$$

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Also consider $$f\left( x \right) =0.$$

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This is $f(x)=Ce^x$ where $C=0$. –  Jacob Mayle Jul 16 at 19:02
@JacobMayle yes. –  newzad Jul 16 at 19:10
$f(x)=0$ is the same as $f(x)=0 \cdot x$ is the same as $f(x) = 0 \cdot e^x$. They are all the same function. –  Jacob Mayle Jul 16 at 19:11
I think this is the best answer. Simple, and to the point. You don't have to think about anything. $e^x$ be darned! –  user1729 Jul 16 at 19:25

This is an answer to the question as it was before.

How to prove that $f(x)=f'(x)$ is false?


You can prove that that statement is false in many ways, but it requires using additional information about who is $f$. For example:

If we are looking at non-zero polynomials. We can show that $f=f'$ is false by looking at the degree. The derivative always decreases the degree of a non-zero polynomial.

More generally, if we are looking at a function that is meromorphic near a point, and has a zero or a pole of positive order, then we can show that $f=f'$ is false by looking at the order of that zero or pole. The derivative always is going to decrease the order of a zero of positive order, or increase the order of a pole of positive order.

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