Applications of graphical transformations in abstract functions

I. Introduction of the issue

When we learn the properties of functions in the first year of high school, we will encounter some problems with abstract functions, so let's look at two examples:
[Example 1]known function$f\left(2x+1\right)$The domain of definition of$\left[1,2\right]$then the function$f\left(4x+1\right)$The domain of definition of .

[Example 2]known function$f\left(x\right)$The domain of definition of$\mathrm{R}$and$f\left(2x-1\right)$The graph of the line about the line$x=1$Symmetry.$f\left(3x+2\right)$is an odd function, then the value of one of the following options must be$0$The ( )
A．$f\left(\frac{7}{2}\right)$ B．$f\left(2024\right)$ C．$f\left(1\right)$ D．$f\left(\frac{3}{2}\right)$
The above two questions do not tell us the analytic form of the function, we can use the method of transforming the graph of the function to deal with similar problems.

II Image Transformation

Image transformations include translation transformations, stretching transformations, symmetry transformations, and flipping transformations, and we only need to understand the first two transformations.
(We'll have a better understanding of trigonometric functions when we've taken Compulsory Book 1, Chapter 5)
Recall, we just learned about primary functions$y=kx+b(k\neq0)$When, how is it known that the graph is a straight line?
It's "list-trace-dot-connect"!

1 Translational transformations

We know the translation mnemonic: add left, subtract right, add up, subtract down, and
Eg1：$y=f(x)$pan left$1$units, pan down$2$The function after one unit is$y=f(x+1)-2$；
Eg2：$y=f(2x-4)$pan left$1$The function after one unit is$y=f[2(x+1)-4]=f(2x-2)$；
Why not?$y=f(2x+1-4)=f(2x-3)$？
Understanding by tracing the dots (less rigorously), the
Angle 1
suppose that...$y=f\left(2x-4\right)$be$y=2x-4$Its graph is translated to the left$1$After one unit, as shown below it is clear that the translationally transformed function is$y=2x-2$.

Angle 2
checkpoint$(x_0,y_0)$exist$y=f(2x-4)$on the graph, then$y_0=f(2x_0-4)$agree on a certain point$(x_0,y_0)$Translated "to the left".$1$After "units" comes$(x_0-1,y_0)$The point is obvious.$(x_0-1,y_0)$exist$y=f(2x-2)$on the image instead of$y=f(2x-3)$.
In short, "left plus and right minus" is a way for$x$Additions and subtractions are made.$y=f[2(x+1)-4]$The parentheses cannot be dispensed with.

2 Scaling transformations

$y=f(x)$The image of the graph has to be telescoped and transformed: the longitudinal coordinate$y$Invariant horizontal coordinate$x$escalate$2$times, i.e.$(x,y)$turn into$(2x,y)$, what would be its analytic formula?
Feel it with concrete examples.
$y=x^2$After telescopic transformation: the vertical coordinate$y$Invariant horizontal coordinate$x$escalate$2$double, point$A(1,1)$up to a point$B(2,1)$；
point (in space or time)$(x_0,y_0)$exist$y=x^2$on the graph, then$y_0={x_0}^2$The points corresponding to the transformed$(x_1,y_1)$Yes$\left\{\begin{array}{l} x_1=2 x_0 \\ y_1=y_0 \end{array}\right.$，
imitate$\left\{\begin{array}{l} x_0=\dfrac{x_1}{2} \\ y_0=y_1 \end{array}\right.$So.$y_1={\left(\frac{x_1}{2}\right)}^2=\frac{x_1^2}{4}$Therefore, the analytic formula of the transformed function is$y=\frac{x^2}{4}$.

as$y=x^2$After telescopic transformation: the vertical coordinate$y$Invariant horizontal coordinate$x$Shrinking to the original$\frac{1}{2}$checkpoint$(1,1)$up to a point$(2,1)$；
The analytic equation of the transformed function is$y={(2x)}^2=4x^2$.

In short, it will$y=f(x)$The horizontal coordinates of the points on the graph of the elongated$(0<b<1)$or shorten$(b>1)$To the original$\frac{1}{b}$times, while the vertical coordinate remains unchanged, to obtain the function$y=f(bx)$The image of .
if$y=f(x)$The vertical coordinates of the points on the graph of the elongated$(a>1)$or shorten$(0<a<1)$To the original$a$times, while the horizontal coordinates remain unchanged, to obtain the function$y=af(x)$The image of .

III Inverse use of graphical transformations

We know the transformation of the analytic formula of a function after the graph of the function has been transformed by a translation transformation or a telescoping transformation.
function (math.)$y=f(2x-2)$How to transform into a function$y=f(x)$And?
Method 1 Panning then telescoping
$y=f(2x-2)=f[2(x-1)]$pan left$1$One unit was given$y=f(2x)$and then the horizontal coordinate$x$escalate$2$obtain$y=f(x)$.
Method 2 Expansion then translation
$y=f(2x-2)$horizontal coordinate$x$escalate$2$obtain$y=f(x-2)$and then pan to the left.$2$One unit was given$y=f(x)$.

IV. Solution of the problem

Having understood the above two points, we now try to find a solution to the two questions posed at the outset.
[Example 1]known function$f\left(\frac{x}{2}+1\right)$The domain of definition of$\left[-1,2\right]$then the function$f\left(2x+1\right)$The domain of definition of .
acrobatic display (esp. on horseback) (old)Using graphical transformations to solve for the
function (math.)$f\left(\frac{x}{2}+1\right)=f\left[\frac{1}{2}\left(x+2\right)\right]$The domain of definition of$\left[-1,2\right]$
$\stackrel {shift 2 units to the right}{\Longrightarrow} $$f\left(\frac{x}{2}\right)$the domain of definition is $[1,4]$
$\stackrel {Transverse coordinates reduced to original \frac{1}{4}}{\Longrightarrow} $$f\left(2x\right)$the domain of definition is $\left[\frac{1}{4},1\right]$
$\stackrel {shift left \frac{1}{2}units}{\Longrightarrow} $$ f\left(2x\right)$the domain of definition is $\left[-\frac{1}{4},\frac{1}{2}\right]$
So the answer is$\left[-\frac{1}{4},\frac{1}{2}\right]$.

[Example 2]known function$f\left(x\right)$The domain of definition of$\mathrm{R}$and$f\left(2x-1\right)$The graph of the line about the line$x=1$Symmetry.$f\left(3x+2\right)$is an odd function, then the value of one of the following options must be$0$The ( )
A．$f\left(\frac{7}{2}\right)$ B．$f\left(2024\right)$ C．$f\left(1\right)$ D．$f\left(\frac{3}{2}\right)$
acrobatic display (esp. on horseback) (old) By the graph of the line about the line$x=1$asymmetrical
$\stackrel {shift left \frac{1}{2}units}{\Longrightarrow} $$f\left(2x\right)$the graph of which is about a straight line$x=\frac{1}{2}$symmetrical) $\stackrel {Horizontal Coordinate Expanded 2x}{\Longrightarrow} $$ f\left(x\right)$the graph of which is about a straight line$x=1$-symmetric

leave it (to sb)$f\left(3x+2\right)$is an odd function with respect to$(0,0)$asymmetrical
$\stackrel {shift right \frac{2}{3}units}{\Longrightarrow} $$f\left(3x\right)$the image about $\left(\frac{2}{3},0\right)$symmetrical) $\stackrel {The horizontal coordinate is expanded 3 times}{\Longrightarrow} $$f\left(x\right)$the image about $($2,0$)symmetric

imitate$f\left(x\right)$The graph of the line about the line$x=1$symmetric and with respect to$(2,0)$Symmetry.
Suppose that when$0<x<1$when$y=x$, repeated use of the above two points, combined with the graph leads to the function$f\left(x\right)$periodicity$4$.

imitate$f\left(2024\right)=f(0)=0$.

$f\left(\frac{7}{2}\right)=f\left(-\frac{1}{2}\right)=-f\left(\frac{3}{2}\right)$，$f\left(1\right)$Neither is sure if the$0$．
So choose:$B$.

Alternative solutions to the problem
[Example 1]This is because the function$f\left(\frac{x}{2}+1\right)$The domain of definition of$\left[-1,2\right]$namely$x\in\left[-1,2\right]$So.$\frac{x}{2}+1\in\left[\frac{1}{2},2\right]$，
leave it (to sb)$\frac{1}{2}\le2x+1\le2$solve for$-\frac{1}{4}\le x\le\frac{1}{2}$，

so the function$f\left(4x+1\right)$The domain of definition of$\left[-\frac{1}{4},\frac{1}{2}\right]$.

[Example 2]$f\left(2x-1\right)$The graph of the line about the line$x=1$Symmetry.
imitate$f\left(2x-1\right)=f\left(2(-x+2)-1\right)=f(-2x+3)$.
assume (office)$f\left(2x-1\right)=f\left(2(-x+2)-1\right)=f(-(2x-1)+2)$，
honorific title$t=2x-1$follow$f\left(t\right)=f(-t+2)$，
imitate$f\left(x\right)$with respect to$x=1$Symmetry.
$f\left(3x+2\right)$is an odd function of
imitate$f\left(3x+2\right)+f(-3x+2)=0$，$f\left(3x+2\right)+f(-(3x+2)+4)=0$，
honorific title$t=3x+2$follow$f\left(t\right)+f(-t+4)=0$follow$f\left(x\right)$with respect to$(2,0)$Symmetry.
moreover$t=2$obtain$f(2)=0$.
From the preceding$f\left(t\right)=f(-t+2)=-f(-t+4)$and such that$t=0$follow$f(0)=f(2)=f(4)=0$.
both (... and...)$f(-t+2)=-f(-t+4)=-(-f(-t+6))=f(-t+6)$，
honorific title$m=-t+2$follow$f(m)=f(m+4)$，
happening$f\left(x\right)$periodicity$4$.
imitate$f\left(2024\right)=f(0)=0$.

$f\left(\frac{7}{2}\right)=f\left(-\frac{1}{2}\right)=-f\left(\frac{3}{2}\right)$，$f\left(1\right)$, neither of which is sure if it's$0$．
So choose:$B$.

V. Consolidation exercises

known function$f\left(2x-1\right)$The domain of definition of$\left(-1,9\right)$then the function$f\left(3x+1\right)$The domain of definition of ( )
A．$\left(-\frac{1}{3},\frac{4}{3}\right)$ B．$\left(-\frac{4}{3},\frac{16}{3}\right)$ C．$\left(-\frac{2}{3},\frac{8}{3}\right)$ D．$\left(-2,28\right)$
known function$f\left(x\right)$arbitrary$x\in R$all have$f\left(x+8\right)=-f\left(x\right)$if$y=f\left(x+2\right)$of the graph about the point$\left(-2,0\right)$symmetric, and$f\left(3\right)=3$follow$f\left(43\right)=$（）
A．$0$ B．$-3$ C．$3$ D．$4$
known function$y=f\left(2x-1\right)$of the graph about the point$\left(-1,1\right)$symmetrically, then the following functions are odd functions ( )
A．$y=f\left(2x-2\right)+1$ B．$y=f\left(2x-3\right)+1$
C．$y=f\left(2x-2\right)-1$ D．$y=f\left(2x-3\right)-1$
known (science)$f\left(x-1\right)$is the domain of definition of$R$The odd function of$g\left(x\right)=f\left(2x+3\right)$is an even function with domain R. Then ( )
A．$g\left(2\right)=0$ B．$g\left(3\right)=0$ C．$f\left(3\right)=0$ D．$f\left(5\right)=0$
(Multiple choice) A known function$f(x)$The domain of definition of$\mathrm{R}$，$f(x+1)$is an odd function of$f(x+2)$is an even function and$x\in[0,1]$when$f(x)$monotonically increasing, then the following conclusion is correct ( )
A．$f(x)$is an even function B.$f(x)$of the graph about the point$(-1,0)$center symmetry
C．$f(2024)=0 $ D．$f\left(\frac{5}{4}\right)+f\left(-\frac{1}{4}\right)<0$

reference answer

[Detailed] Functions$f\left(2x-1\right)$The domain of definition of$\left(-1,9\right)$，
$\stackrel {shift left \frac{1}{2}units}{\Longrightarrow}$function$f\left(2x\right)$the domain of definition is $\left(-\frac{3}{2},\frac{17}{2}\right)$， $\stackrel {The horizontal coordinate is reduced to the original \frac{2}{3}times}{\Longrightarrow}$function$f\left(3x\right)$the domain of definition is $\left(-1,\frac{17}{3}\right)$， $\stackrel {shift left \frac{1}{3}units}{\Longrightarrow}$function$f\left(3x+1\right)$the domain of definition is $\left(-\frac{4}{3},\frac{16}{3}\right)$. Hence: $B$.
[Detailed Explanation] As$y=f\left(x+2\right)$of the graph about the point$(-2，0)$Symmetry.
happening$y=f\left(x\right)$of the graph about the point$(0，0)$symmetry, i.e.$y=f\left(x\right)$is an odd function of
both... and...$f\left(x+8\right)=-f\left(x\right)$，
imitate$f\left(x+16\right)=-f\left(x+8\right)=f\left(x\right)$That's $16.$for $f\left(x\right)$the period. Let $x=-3$Substitute $f\left(x+8\right)=-f\left(x\right)$, then $f\left(5\right)=-f\left(-3\right)=f\left(3\right)=3$. \$f\left(43\right)=f\left(43-3\times16\right)=f\left(-5\right)=-f\left(5\right)=-3$. Therefore: $B$
[Detailed] Functions$y=f\left(2x-1\right)$of the graph about the point$\left(-1,1\right)$Symmetry.
so the function$y=f\left(2x-1\right)$The graph of$1$Units.
The graph of the function after translating down one unit about the point$\left(0,0\right)$Symmetry.
(i.e. can be obtained by pressing a button)$y=f\left[2\left(x-1\right)-1\right]-1=f\left(2x-3\right)-1$.
So choose:$D$
[Details] Because$f\left(x-1\right)$is the domain of definition of$R$The odd function of
so the function$f\left(x\right)$About the point$\left(-1,0\right)$symmetric and $f\left(-1\right) = 0 $
on account of$g\left(x\right)=f\left(2x+3\right)$is the domain of definition of$R$The even function of
so the function$f\left(x\right)$About Straight Line$x=3$Symmetry.
the reason why$f\left(7\right)=0$namely$g\left(2\right)=0$.
So choose:$A$
[Details] Because$f(x+1)$is an odd function, so$f(x)$with respect to$(1,0)$Symmetry ①.
on account of$f(x+2)$is an even function, so$f(x)$with respect to$x=2$Symmetry ②.
$x\in[0,1]$when$f(x)$monotonically increasing.
Suppose that when$0\le x\le1$when$y=x-1$，
Repeatedly using ① and ② above, the
Combining the graphs gives the function period$T=4$，$f(x)$is an even function.$f(x)$of the graph about the point$(-1,0)$Center symmetry.$f(2024)=f(0)$Not necessarily$0$.

happening$AB$Correct.$C$Error;
on account of$f\left(\frac{5}{4}\right)+f\left(-\frac{1}{4}\right)=f\left(2-\frac{3}{4}\right)+f\left(\frac{1}{4}\right)$$=-f\left(-\frac{3}{4}\right)+f\left(\frac{1}{4}\right)=f\left(\frac{1}{4}\right)-f\left(\frac{3}{4}\right)$，
both (... and...)$x\in[0,1]$when$f(x)$monotonically increasing.
the reason why$f\left(\frac{1}{4}\right)<f\left(\frac{3}{4}\right)$namely$f\left(\frac{1}{4}\right)-f\left(\frac{3}{4}\right)<0$，
happening$D$Correct.
So choose:$ABD$.