I Can Two Different Functions Have the Same Output for a Single Point?

[kent davidge]

Today, while reading about bijections, a question came into my mind. And that is: is there any way that two different functions $f$ and $g$ acting on a same point $p$ gives the same output? In symbols, as I'm not good in English, is it possible that $f (p) = g(p)$ with $f \neq g$?

 

[kent davidge]

Oh and perhaps I should say I'm considering the same domain for the two functions.

 

[fresh_42]

Today, while reading about bijections, a question came into my mind. And that is: is there any way that two different functions $f$ and $g$ acting on a same point $p$ gives the same output? In symbols, as I'm not good in English, is it possible that $f (p) = g(p)$ with $f \neq g$?

This happens all the time: $p=0$ for $f(x)=x^2\; , \;g(x)=\sin(x)\; , \;h(x)=|x|$ etc. However $f=g$ if $f(p)=g(p)$ for all points $p \in \operatorname{dom}(f)=\operatorname{dom}(g)$. We also widely use the fact, that functions are the same at one point and close in their neighborhood, when we approximate a function by its tangent in $p$.

 

[Mark44]

Today, while reading about bijections, a question came into my mind. And that is: is there any way that two different functions $f$ and $g$ acting on a same point $p$ gives the same output? In symbols, as I'm not good in English, is it possible that $f (p) = g(p)$ with $f \neq g$?

Here's an example that's almost what you're talking about:
$f(x) = \frac{x^2 - 1}{x - 1}$ and $g(x) = x + 1$
f(x) = g(x) everywhere except at x = 1, the only point where f is not defined. Although the two functions have the same values almost everywhere, the domains are different (the domain of f doesn't include x = 1, and the domain of g is all real numbers), so they are different functions.