# Function Transformation Graph

Consider a real function $g(t)$. Now consider another real function $f(t) = -t +3$ that transforms the domain of $g(t)$. Suppose I have the graph of $g(t)$ and I'm trying to plot the $g(f(t))$ graph by drawing. In order to get the correct graph, I could think of $f(t) = -(t-3)$, in which case, to plot $g(f(t))$, I should first invert $g(t)$ with respect to the vertical axis and then shift $3$ units to the right.The problem is that I'm not seeing the intuition into inverting $t$ first, and then shifting the inverted graph by 3 units to the right. Why is thinking of shifting 3 units to the right first $(t-3)$ and then inverting the shifted graph $-(t-3)$, a wrong procedure? Why is thinking of inverting the domain $( -t )$ and then shifting the inverted graph 3 units to the left $( -t + 3 )$ a wrong procedure ?

I got the intuitive feeling for all transformations, both of domain and of the range.This one though, is where I fall into confusion. Thanks

Welcome to Math.SE! Sorry there's been a little delay with answering your question.

A function is a sequence of operations. The composition $g(f(t))$ could be described as a sequence of the following operations:

(a) Subtract $3$ from $t$; (b) Change the sign of the result; (c) Apply $g$.

Both of these are valid ways to think about what the composition does. When constructing a graph, we begin with $g$ (since this is what we have already), which is step (c) in either case. Since (c) is understood already, next thing is to understand what happens if we do (b) and then (c). Notice that it would be wrong to think of "(a) and then (c)", because this is not the order in which things are done in the construction of $g(f(t))$.

So, this is how we are led to introduce "invert in vertical axis", step (b). Finally it is time to bring (a) into the fold, which adds the shift by 3 units to the right.

Although I don't want to contribute to confusion, it seems natural to mention that the same composition $g(f(t))$ can be also described as a different process:

(a) Change the sign of $t$; (b) Add $3$ to $t$; (c) Apply $g$.

If you interpret these as manipulations on graph of $g$, it will be "shift the graph to the left (b); and then invert about vertical axis (a)." Sounds different, but gives the same result.

Let me just illustrate Pavel's excellent answer with an example. To use a somewhat visually interesting example, suppose our function is $$g(t)=2\sin(t)+t$$ which has the following graph from $t=-6$ to $t=6$: If we think of $f(t)$ as first translating (i.e., "sliding") $t$ over by 3, then negating the result, the graph of $g$ applied to $f(t)$ as we make this transformation looks like this: If instead we think of $f(t)$ as negating $t$, and then translating the result over by 3, the graph of $g$ applied to $f(t)$ as we make this transformation looks like this: As you can see, we get the same result either way.