Student. -Student: If the function g$g$ is such that g(x)= f(x-1)$g(x)= f(x-1)$, should not the graph of g be ,$g$ be at every point, 1 unit to the left of the graph of f$f$?
Teacher. -Teacher: Why do you think it should be the case?
Student. -Student: Well I see that in the equation g(x)$g(x)$ ( onon the left) becomes f(x-1)$f(x-1)$ ( onon the right). Apparently, the function symbol "f""$f$" is substituted for the function symbol " g"$g$" and the variable " x-1 "$x-1$" is substituted for the variable "x""$x$". So the X-coordinate of each " new"new point" decreases ofby 1 unit , and the whole graph is moved 1 unit to the left.
Teacher.-Teacher: Let me ask you a question: which variable represents, according to you, the input of the function g$g$?
Student -Student: Normally it should be the variable x$x$, since the left side of the equation has " g(x)$g(x)$", but the whole equation says that instead of x$x$, the input becomes (x-1)$(x-1)$. In the expression " f (x-1)$f(x-1)$", does not the number (x-1)$(x-1)$ play the role of the input?
Teacher -Teacher: I think your mistake as to the direction of the transmormationtransformation comes precisely from not understanding that the inputs that g takes are represented by the letter " x"$x$". The number (x-1)$(x-1)$ does not play the role of input ( for gfor $g$) ;; it is simply "used" by the function-machine g$g$ ( soso to say) to calculate the image of x$x$ under g$g$.
Student -Student: Could you takeshow an example.?
Teacher. -Teacher: Let me take a general example. supposeSuppose that a point ( x, y)$(x, y)$ belongs to the graph of g$g$. Knowing that y$y$ is g(x)$g(x)$ and that g(x) = f(x-1)$g(x) = f(x-1)$, how could you rewrite the couple defining this point?
Student.-Student: Probably ( x-1, f(x-1) )$(x-1, f(x-1))$, no?
Teacher -Teacher: Why did you change the X-cordinatecoordinate of this point? You've "moved" that point to the left without any reason! The only new information I gave you is about " y"$y$" , not about " x"$x$".
Student-Student: You're right! Using the information you gave me, I'll now say that
(x, y) = ( x, (f(x-1))
$$(x, y) = (x, (f(x-1))$$
Teacher .-Teacher: Right! Now, tell me, which point belonging to the graph of f corrrespondscorresponds to our " g$g$-point" : ( x, f(x-1)) $(x, f(x-1))$? Obviously, the Y-coordinate of that point is f(x-1)$f(x-1)$, but what is its X-coordinate?
Student.-Student: The answer is in the question itself, isn't it? It's obviously the point
(x-1, f(x-1) ) !
$$(x-1, f(x-1))$$
Teacher.-Teacher: I agree! But do you realize what it means? You've just told me that for any arbitrary point (x, y)$(x, y)$ belonging to the graph of g$g$, the corresponding point belonging to the graph of f$f$ has the same Y-cordinatecoordinate, but an X-coordinate that is smaller by 1 unit! That means that any arbitrary " g$g$-point" ( soso to say) is one unit to the RIGHT of it's corresponding point "f"$f$-point"!
Student.Student: If I now understand what I said myself, I've just claimed, contrary to what I contended at the beginning, that the transformation is a translation of the whole graph of f$f$ 1 unit to the RIGHT!
Teacher.-Teacher: That's what you said, actually!
