How to parameterize the maximum of a function?

Question

$f(x) := -4(\frac{1}{2}-x)^2+1$

Is it possible to construct a parameterized version of $f$, say $f_a$, which fulfills:

• $f_a(0) = f_a(1) = 0$
• $f_a(a) = 1$
• $f_a'(x) = 0 \Leftrightarrow x=a$
• $f_a''(a) < 0$
• $f_{1/2} = f$
• strictly monotonically increasing on [0,a)
• strictly monotonically decreasing on (a,1]
• $f_a$ does not have to be a polynomial necessarily
• $f_a$ is differentiable on [0,1]

I think you get the idea - just imagine a function like $f$ and visually shift the maximum along the red line while keeping anything else as (subjectively) constant as possible.

I would also be interested in learning whether this can be done numerically or if there is a systematic analytical method to achieve something like this. So I could apply this method to more complex functions. Also a definite "this is not possible because ..." would be a welcome answer.

Answer 1

Let $f_a(x) = c_1x^2 + c_2x + c_3$. Since $f_a(0) = 0$, $c_3 = 0$. Since $f_a(1) = 0$, $c_1 = -c_2$. Therefore, $$f_a(x) = c_1(x^2 - x)$$

Since $f_a(a) = 1$, therefore $c_1(a^2 - a) = 1$, i.e. $c_1 = \frac{1}{a^2 - a}$. Therefore, $$f_a(x) = \frac{1}{a^2 - a}(x^2 - x)$$

Now, $f_a'(x) = \frac{1}{a^2 - a}(2x - 1)$. For $f_a'(a) = 0$, $a$ has to be equal to 1/2. Doesn't look there is any other choice :-/

Answer 2

$$f(x)=\begin{cases}1-(x-a)^2/a^2, x<a\\1-(x-a)^2/(1-a)^2, x\ge a\end{cases}$$

It is differentiable, but unfortunately it has no second derivative at $x=a$.