# How do I find a function based on given function values?

Let us suppose I have an unknown function $$f(x)$$.

I have the information that $$f(2)=6$$ and $$f'(2)=-1$$.

What would the actual function be? Are there unlimited possibilities? How do I make an expression that satisfies this information?

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• You cannot tell what the whole function is by the information you are providing. The simplest way to find one(!) function is to chose $a,b$ properly such that $f(x)=ax+b$ satisfies your conditions.
– user592521
Sep 10, 2021 at 9:44
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Sep 10, 2021 at 9:51

There are multiple fonctions that satisfy this. The "easiest" way is to consider a polynomial function as $$f\left(x\right)=ax+b$$

You have $$f\left(2\right)=2a+b=6$$ and $$f'\left(2\right)=a=-1$$ Hence an example is $$f\left(x\right)=8-x$$

Yes: there are unlimited options. One of the simplest would be the affine function

$$f(x)=-x+8$$ which goes through the point $$(2,6)$$ and whose slope is $$-1$$ (actually everywhere), as requested.

As it was mentioned in other answers, there are countless possibilities. To obtain a relatively large selection of such functions, you can draw inspiration from Taylor's series, where $$f(x) = \sum_{n=0}^{+\infty}\frac{f^{(n)}(a)}{n!} (x-a)^n, \quad x \in D.$$

Any function of the form $$f(x) = 6 - (x-2) + \sum_{n=2}^{+\infty} c_n (x-2)^n$$

is a good fit, as long as the series is convergent. In particular, taking $$c_n = 0$$, you get $$f(x)=6-(x-2)$$.