I Transformation of Functions: How Do Domain and Range Change?

AI Thread Summary
The discussion centers on how transformations affect the domain and range of the function f(x)=2-x defined on [0,2]. Transformations such as f(-x), -f(x), f(x+3), and others are analyzed, revealing how each modifies the domain and range while maintaining the underlying rule of f. For instance, f(-x) changes the domain to [-2,0], while f(x+3) shifts the domain to [-3,-1]. The participants clarify that despite these transformations, the fundamental rule of f remains unchanged. Overall, the thread emphasizes understanding the impact of various transformations on functions without altering the original function's definition.
DumpmeAdrenaline
Messages
80
Reaction score
2
I want to understand how the domain and range change upon applying transformations like (left/right shifts, up/down shifts, and vertical/horizontal stretching/compression) on functions.
Let f(x)=2-x if 0 ≤x ≤2 and 0 otherwise.
I want to describe the following functions 1) f(-x) 2) -f(x) 3) f(x+3) 4) f(x)+3 5) f(2x) 6) 2f(x) 7) f(2x+3)
The rule generates an output by multiplying the input by -1 and adding the result to 2.
1) f operates on the closed interval [0,2] as its domain. With f(-x) the domain then changes to [-2,0] so that upon multiplying each real number in this interval by -1 we obtain the same domain [0,2] and the same image [0,2] as f(x). f(-x)=2-(-x)=2+x.
2) -f(x) change the sign of the output keeping the input.
3) f(x+3) Each input in the closed interval [0,2] moves by 3 units, such that the domain changes to 3 ≤x+3 ≤5. However, under this domain the function generates an output of 0. As a result the domain of f(x+3) becomes [0,-1] so that upon adding 3 units to we wind up with the same domain and range as f(x).
4) Move every point up by 3 units
5) The domain of f(2x) shrinks to 0<x<1 so that when multiply each real number we obtain the same domain and range as f(x). f(2x)=2-2x
6) Multiply each output by 2 keeping the input.
7) Composed transformation of shrinking by 1/2 followed by a shift 3 units to the left resulting in a domain of [-3,-1].
In all of the above are we changing the rule f?
 
Mathematics news on Phys.org
DumpmeAdrenaline said:
I want to describe
You also want to learn how to post math using a little ##\ \LaTeX\ ##.
See tutorial. It's really easy, and it's fun. Enclose in ## for in-line math and in $$ for displayed math. Example:
Code:
$$1) \quad  f(-x) \\ 2) \quad -f(x) \\ 3) \quad f(x+3) \\
4) \quad f(x)+3 \\5) \quad f(2x) \\6) \quad 2f(x) \\7) \quad f(2x+3)$$
yields
$$1) \quad f(-x) \\ 2) \quad -f(x) \\ 3) \quad f(x+3) \\4) \quad f(x)+3 \\5) \quad f(2x) \\6) \quad 2f(x) \\7) \quad f(2x+3)$$

[edit] this looks like ... because MathJax doesn't acknowledge the line feeds (\\\) and I don't understand what is screwing things up. :cry:

A better way to do this is with \begin{align} and let ##\TeX## do the numbering:
Code:
$$\begin{align}
 & f(-x)  \\  -&f(x) \\ &f(x+3) \\ &f(x)+3 \\&f(2x) \\2&f(x) \\ &f(2x+3)
\end{align}$$
$$\begin{align}
& f(-x) \\ -&f(x) \\ &f(x+3) \\ &f(x)+3 \\&f(2x) \\2&f(x) \\ &f(2x+3)
\end{align}$$

##\ ##
 
Last edited:
DumpmeAdrenaline said:
3) f(x+3) Each input in the closed interval [0,2] moves by 3 units, such that the domain changes to 3 ≤x+3 ≤5.
No. Make a sketch to see. You doubled up: domain is ##0 \le x+3 \le 2## .

DumpmeAdrenaline said:
In all of the above are we changing the rule f?
No, we are not.

##\ ##
 
BvU said:
No. Make a sketch to see. You doubled up: domain is
I reached the same answer but for the wrong reason.I thought that the x in f(x) is the same as x+3. Hence, why I added 3 to both sides of the inequality. So, we regard x+3 and x as just labels to the input?
 
From post #1:
DumpmeAdrenaline said:
3) f(x+3) Each input in the closed interval [0,2] moves by 3 units, such that the domain changes to 3 ≤x+3 ≤5. However, under this domain the function generates an output of 0. As a result the domain of f(x+3) becomes [0,-1] so that upon adding 3 units to we wind up with the same domain and range as f(x).

BvU said:
No. Make a sketch to see. You doubled up: domain is ##0 \le x+3 \le 2## .
And further, ##0 \le x+3 \le 2 \Rightarrow -3 \le x \le -1##, so the domain for y = f(x + 3), is the translation left by 3 units of the domain for y = f(x).
DumpmeAdrenaline said:
I reached the same answer but for the wrong reason.I thought that the x in f(x) is the same as x+3.
You're dealing with two different functions, each of which is a translation by 3 units of the other.

Here's a sketch of y = f(x) (in blue) and y = f(x + 3) (in red).
graph.png
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top