Looking for a function with finite support for which $f' = f$ with finite support Does there exist a differentiable function with the following properties:
$f(0) = 1$
$0 < f(x) < 1$ for $ 0 < x < 1$
$f(x) = 0$ for $x \geq 1$
$f'(0) = 0$
and lastly, $f'(p) = f(p)$ for at least one value of $p$ between $0$ and $1$.
 A: The function equal to $(x-1)^2(2x+1)$ for $x\leq 1$ and zero elsewhere satisfies all conditions except the last one ($p=1$ satisfies the condition, but no $p$ between $0$ and $1$).
However, you can modify this function by adding a continuously differentiable bump function $h(x)$ which is supported on a small neighborhood of any point in $(0,1)$ - for instance, $x=\tfrac12$ - in such a way that the derivative condition is satisfied in a neighborhood of $x$. This is possible because all the other conditions other than the last one are local to the endpoints.
To be a little more concrete, let the bump function $h(x)$ satisfy the following conditions:


*

*$h(\tfrac12)=0$

*$h'(\tfrac12)=\tfrac12$

*$h(x)=0$ if $x\not\in (\tfrac14,\tfrac34)$

*$|h(x)|<\tfrac{5}{32}$ for all $x$
Then $f(x)+h(x)$ satisfies the conditions with $p=\tfrac12$. (Note that $f(\tfrac12)=\tfrac12$ and the $\tfrac{5}{32}$ comes by considering the values of $f(\tfrac14)$ and $f(\tfrac34)$ (to ensure that $f(x)+h(x)$ remains in $(0,1)$.
A: Define $f(x) = \begin{cases} {1 \over 2} (54 x^3 -27 x^2 +2), & x \in [0, {1 \over 3}) \\
{1 \over 2} + {1 \over 8} (1-\cos (24 \pi (x-{1 \over 3}))),& x\in [{1 \over 3}, {2 \over 3})\\
{27 \over 2} (x-1)^2(2x-1), & x \in [{2 \over 3},1] \\
0, & x>1\end{cases}$.
Check that $f(0)=1$, $f'(0) = 0$, $f(x) = 0$ for $x \ge 1$, $f(x) \in (0,1)$ for all $x \in (0,1)$.
Since $[-8,8] \subset f'([{1 \over 3}, {2 \over 3}])$, it is clear that there is some $x \in (0,1)$ such that $f(x) = f'(x)$.
Note the function is much easier to construct than to figure out where the above definition came from. I started with the polynomial $x(x-{1 \over 3})$ and integrated, scaled & shifted. This defined the function on $[0, {1 \over 3}]$. Then I used this to define the 'complementary' function on $[{2 \over 3}, 1]$. Then I filled in the middle with a shifted & scaled $\cos$ to that the derivative would be all over the place.
