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Could somebody please explain the reasons why a function y=f(2x) is stretched horizontally (compared to y=f(x)) and why the function y=f(x+1) is shifted to the left (compared to y=f(x)).

Bobbo
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  • If you find it counter intuitive that $f(x + 1)$ is shifted to the left because it has a plus: if $g(x) = f(x + 1)$, then $g(x - 1) = f(x)$, which makes sense because $\mathbf{g}$ is $f$ shifted to the left. – user388557 Apr 01 '19 at 07:52

3 Answers3

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Given two functions $f, g$, where $g(x):=f(x+1)$, we can say g is a "one-unit-left-shifted" version of f, since anything you see for $f$ at $x=1$, you will find it for $g$ at $x=0$. Similarly, $f(2)=g(1)$ and so on for every real number.

If g was defined so that $g(x):=f(2x)$, that would be the equivalent of "walking through the x axis at twice the speed". For instance, $g(0)=f(0)$, but what you see for $f$ in the interval $[0,1]$, you will see if for $g$ "happening" (meaning, all the same values will be reached) in only $[0, 1/2]$

If you still don't see it clearly, just make a couple of graphs yourself and compare!

David
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Doing this shift $f(x) \to f(x+1)$, you are in fact doing a change of coordinates, which, "technically", works backwards.

For understanding it, let us take the example of the equation :

$$(x-x_0)^2+(y-y_0)^2=1$$

(circle centered in $(x_0;y_0)$ with radius $1$).

For the sake of explanations, let us consider that $x_0$ and $y_0$ are $>0$.

If we refer to the standard equation of the unit circle $X^2+Y^2=1$, it is as if the unit circle was "pushed forward" in the first quadrant ; but the change of coordinates for doing that :

$$\begin{cases}X&=&x-x_0\\Y&=&y-y_0\end{cases}$$

looks "backwards".

It will look a "forward" action if we express the old coordinates as functions of the new ones (which is a general technique) :

$$\begin{cases}x&=&X+x_0\\y&=&Y+y_0\end{cases}$$

Take a look at http://doubleroot.in/lessons/coordinate-geometry-basics/translation-of-axes/ for a similar explanation.

Jean Marie
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Contemplate that if before the transformation, you have

$$f(x)\to (0,f_0),(1,f_1),(2,f_2),(3,f_3),\cdots$$

after the stretching you have the points

$$f(2x)\to (0,f_0),(0.5,f_1),(1,f_2), (1.5,f_3),\cdots$$

If you plot those points, they appear to be closer horizontally.

Now considering the shift, you have

$$f(x+1)\to (-1,f_0),(0,f_1),(1,f_2),(2,f_3),\cdots$$

and the points have moved to the left.