Differentiation
Let f : R → R be a function such that f ( 2 ) = 0 and f ′ ( x ) = f ( x ) . What is f ( 2 0 1 5 ) ?
The answer is 0.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
5 solutions
y = e x − e 2 c a n n o t satisfy d x d y = y a n d f ( 2 ) = 0 .
If this can be the case, then f (2015) = e 2 0 1 5 − e 2 = 1 . 2 6 8 7 6 4 5 4 8 0 8 1 0 2 8 2 4 4 7 3 4 0 9 1 2 7 3 8 0 8 e + 8 7 5 .
Since the answer wanted requested for an integer, y = 0 as a constant can satisfy the conditions.
Where no contradiction is there, y = f (2015) = constant zero = 0 cannot be denied.
Therefore, f (x) is proposed by someone as a e x − b where a has to be zero.
Answer: 0
this function f ( x ) = e x − e 2 has its derivative f ′ ( x ) = e x therefore it does not satisfy the condition that f'(x)=f(x)
Wrong. This solution made no sense.
An easy approach might be by starting with integrating both sides which gives lnf(x)=x+C(some arbitrary constant) which gives f(x)=e^x+c which is always positive. So for f(2)=0 f must identically vanish. So f(x)=0 for all x belong to real set.
The derivative of f ( x ) = a × ( e x − b ) is the same function, where a and b are constants, but this function is never equal 0, except when a = 0 , thus the function must be f ( x ) = 0 . Finally, it implies f ( 2 0 1 5 ) = 0 .
The first line is not true. If it was the only function, then f ( x ) = 0 cannot satisfy the conditions.
Be careful to double check that what you wrote is what you intended. It can be tricky to convey exactly what you intended.
Log in to reply
Can't the function be f(x)=0? then f'(x)=0 and f(2)=0 as well. It was just a tricky one.
Log in to reply
I believe he is saying that "The only function that is equal to its own derivative is f ( x ) = e x " is a false statement.
All functions of the form f ( x ) = A e x also satisfy it.
So since e x = 0 we conclude A = 0 and thus f ( x ) = 0 .
Log in to reply
@Isaac Buckley – Right... Thanks for clarification :)
Log in to reply
@Zeeshan Ali – Yes, it can be tricky writing a complete proof that expresses exactly what we are thinking.
The important part of this problem is showing that " f ′ ( x ) = f ( x ) if and only if f ( x ) = A e x for some A .
Currently, what you wrote is "If f ( x ) = A e x , then f ′ ( x ) = f ( x ) .". As such, it only explains why "For this particular family of solutions, we must have A = 0 ". It doesn't explain "Why aren't there other solutions with a different value of f(2015)"?
Log in to reply
@Calvin Lin – I don't have any other possible outcome for f(2015) still.
I just thought of f (x)=0, it's a linear equation that's a derivative of itself. And also satisfied f (2)=0. It may not be a good solution, but it works. Therefore, if x=2015, f (2015)=0
You can solve Cauchy's problem y'=y with y(2)=0. y is continue (y'=y implies the existence of derivative)
The derivative of f ( x ) e − x is f ′ ( x ) e − x − f ( x ) e − x = 0 , so that f ( x ) e − x = C , a constant. Plugging in x = 2 , we see that C = 0 × e − 2 = 0 . It follows that f ( x ) = 0 × e x = 0 for all x ; in particular, f ( 2 0 1 5 ) = 0 .