Tricks with transforming function plots

Question

I want to try a function graph trick and need some advice.

We have a function graph like this:

The trick is as follows: We take an arbitrary interval $x∈[a;b]$ and the corresponding section of the function graph $f$. Further, we must introduce a new function F, such that for $t_0=0$ its value is equal $Y=y_a$, and for $t_k = b$ its value is equal $Y=y_b$. This can be graphically displayed as follows.

Sorry for the messy design. I was in a hurry. Can such a trick be done?

Answer 1

If i understand you correctly, the only thing you need is a linear parametrization mapping $$t=0 \mapsto b \mbox{ and } t=b \mapsto a $$ $$\Rightarrow x(t) = a + \frac{b-a}{b}(b-t)$$

As you can see: $x(0) = b$ and $x(b) = a$

So, plugging this into your function $f$ gives your desired function

$$F(t) = f\left(a + \frac{b-a}{b}(b-t)\right) \text{ on } [0,b]$$

Additional info after comment:

In order to "freeze" the maximum for $b>0$ and using an analytic expression you can do the following:

Assuming that $F(b) = f(a)$ is the (allowed) maximum you can do the following:

$$H(t) = \frac{F(t) + F(b)}{2} - \frac{|F(t) - F(b)|}{2} \text{ on } [0,+\infty)$$

This cuts off $F$ at $F(b)$ and continues with $F(b)$ for $t>b$, provided that $F$ would otherwise continue above the given threshold.

Answer 2

Let $$ F(x) = \begin{cases} f(a) & x \le 0 \\ f(\frac{b-a}{b}x + a) & 0 \le x \le b\\ f(b) & x \ge b \end{cases} $$ and you've got the function that the text under the right-hand graph indicates, although not what the picture in the right-hand graph indicates.