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Please provide any formula or step-by-step guide on how to generate a function for the following graph.

Graph logic:

Part 1 $\longrightarrow$ where $x \leq 3$ $\longrightarrow$ linear relation.

Part 2 $\longrightarrow$ when $3 \leq x \leq 6$ $\longrightarrow$ $y$ remains unchanged.

Then, again part 1 with $3 < x < 9$ $\longrightarrow$ linear relation.

Then part 2 when $y$ remains unchanged.

Thanks!

EDIT: Values for $x$ and $y$

x y
0 0
1 1
2 2
3 3
4 3
5 3
6 3
7 3
8 3
9 3
10 4
11 5
12 6
13 6
14 6
15 6
16 6
17 6

and so on.

I am trying to create an interpolator for the following case:

Interpolation time 3500 ms
Value start = 0, end = 1

Firstly, the 500 ms value increases by the formula (end/3500)*interpolated time.

Then 1000 ms value remains unchanged.

Then 500 ms again value increases.

Then 1000 ms value remains unchanged.

Then 500 ms again value increases.

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  • $\begingroup$ What is the value of $f(x) = y$ in the interval $[3, 9]$? You have to construct a piecewise function. Define $f(x)$ on separate intervals of $x$, just like you have verbally done in your question. $\endgroup$ Commented Apr 23, 2019 at 15:47
  • $\begingroup$ Do you know how to get the equation of of line if you are given two points on the line? When you ask a question here it helps a lot if you tell us what you have tried and what you know about the question. Otherwise, we don't know where to start with an answer. $\endgroup$
    – saulspatz
    Commented Apr 23, 2019 at 15:48
  • $\begingroup$ @Joshiepillow Thanks, thats was my mistake $\endgroup$
    – Ololoking
    Commented Apr 23, 2019 at 16:01
  • $\begingroup$ @saulspatz I've updated my question with a bit more explanations. Please check it $\endgroup$
    – Ololoking
    Commented Apr 23, 2019 at 16:02
  • $\begingroup$ @JacobCheverie I've updated my question with a bit more explanations. Please check it $\endgroup$
    – Ololoking
    Commented Apr 23, 2019 at 16:03

2 Answers 2

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$y = x - 2a/3$ for $a \leq x < a + 3$

$y = a/3 +3$ for $a+3 \leq x < a + 9$

Where $a$ is defined as all numbers for which $a \bmod 9 = 0$.

Basically, $a$ is divisible by 9.

Tried this on Desmos. It works but you’ll have to manually type in every value of $a$.

enter image description here

enter image description here

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If you don’t want to use discrete functions, you can use sigmoid approximation.

$$f_1(x) = \lim_{k\rightarrow \infty}\frac{1}{1 + e^{-k(x)}} \cdot \frac{1}{1 + e^{k(x-3)}}$$

The above function will return approximately $0$ if $x$ is not in the interval $(0,3)$, otherwise $1$. Likewise you can create your own functions for other intervals.

\begin{align} f_2(x) &= \lim_{k\rightarrow \infty}\frac{1}{1 + e^{-k(x-3)}} \cdot \frac{1}{1 + e^{k(x-9)}} \\ f_3(x) &= \lim_{k\rightarrow \infty}\frac{1}{1 + e^{-k(x-9)}} \cdot \frac{1}{1 + e^{k(x-12)}} \\ f_4(x) &= \lim_{k\rightarrow \infty}\frac{1}{1 + e^{-k(x-12)}} \cdot \frac{1}{1 + e^{k(x-17)}} \end{align}

Now, multiply all the functions with their lines.

$$f(x) = f_1 \cdot x + f_2 \cdot 3 + f_3 \cdot (x-6) + f_4 \cdot 6$$

Here is an example MATLAB code and the result graph.

Matlab code and graph

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