My Derivative Is Myself
How many functions are there such that f ( x ) = f ′ ( x ) ?
Clarification: While functions such as f ( x ) = 2 x and f ( x ) = 2 x + 1 are in the same family of functions, they are different functions, so if you found those two as the only answers, your answer would be 2.
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3 solutions
f ( x ) = C e x + k ?
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f
(
x
)
=
C
e
x
+
k
=
(
C
e
k
)
e
x
=
A
e
x
They are basically the same thing with different values of the constant.
If we use the differential equation approach:
(let
y
=
f
(
x
)
)
y
=
d
x
d
y
d
x
=
y
d
y
∫
d
x
=
∫
y
d
y
x
+
C
′
=
ln
y
y
=
e
x
+
C
′
=
e
C
′
e
x
=
C
e
x
So are of them are in the from
C
e
x
f(x)=0=f'(x)
Hint for the bonus: assume f ( x ) = f ′ ( x ) . What is the derivative of e − x f ( x ) ?
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d
x
d
e
−
x
f
(
x
)
=
−
e
−
x
f
(
x
)
+
e
−
x
f
′
(
x
)
=
e
−
x
(
f
′
(
x
)
−
f
(
x
)
)
=
0
So
e
−
x
f
(
x
)
is a constant.
So
f
(
x
)
=
C
e
x
for some constant
C
( 0 x + n ) ′ = 0
We have n different functions, as n is integer and x + n = 0
Using diffrential equations. f ( x ) = e ( x + c )
d x d e x = e x .
By the derivative of a function times a constant rule, we get d x d C e x = C e x . Since C is a real number, we can choose any C to get f ( x ) = C e x as a solution to this property, thus there are infinitely many solutions.
Bonus: Are there any functions not in the form C e x that are their own derivative?