Which function is not one-to-one?

  • Thread starter Thread starter UrbanXrisis
  • Start date Start date
  • Tags Tags
    Function
AI Thread Summary
The discussion centers on identifying which functions are not one-to-one among the given examples F(X), G(X), and H(X). G(X) is determined to be not one-to-one due to repeated values, failing the Horizontal Line Test. F(X) is considered one-to-one if viewed as discrete points, but if imagined as a continuous graph, it would not be. The conversation also touches on the characteristics of odd functions, noting that functions symmetrical about the origin meet this criterion. Overall, the key takeaway is that a function is not one-to-one if it has duplicate output values.
UrbanXrisis
Messages
1,192
Reaction score
1
There is a Math problem that I got wrong and I want to know why it’s wrong:

X...(–2)..(–1)..0...1...2
F(X) 1...(–2)..0.(–1)..2
G(X) 3...(–2)..0.(–2)..3
H(X) 1...(–3)..0...3.(–1)

Which function is not one-to-one? I put all of them but got it wrong
Which function is symmetrical about the origin? is it H(X)?

how would I know if something is one-to-one without graphing it?
What about if a function is symmetrical about the origin?
 
Last edited:
Physics news on Phys.org
You know if it is one-to-one if there is no duplicate value of "y".

Remember, the horizontal test? If it hits the graph twice when you test it as y=3, than there will be two solutions for x, when y=3. You do not want this.

Also, a function that is symmetrical about the origin is an odd function. An odd function is when f(-x)=-f(x).

I can barely tell what you wrote down. I don't see the significance of all the dots.

I am assuming that the top row is the values of x (domain), and the next three rows below are the range of the various functions.

By looking at F(x), we see that we none of the F(x) values are equal. Assuming that the function is just dots, than this function is one-to-one. If we must imagine a line being drawn from dot to dot, than it is NOT one-to-one.

EXPLAINED: We start at point (-2,1), then we go down to point (-1,-2) and then back up to (0,0). To go down then back up, we must have obviously intersected an horizontal line twice at any value F(x).

Try it. Draw a graph that goes up and down. Does it pass the Horizontal Line Test? Now, that draw the same graph of F(x) with no lines, and just the points given. Does this pass the Horizontal Line Test?

Of course, the question seems vague, but it also seems like you might have left something out.

Moving down the list of functions. G(x) has value -2 twice. Is this a one-to-one function? Of course not. Whether you draw the graph with lines or just points, it will fail the Horizontal Line Test.

Now, let's find the odd function. I am assuming you know what an odd function looks like on a graph. If not, graph the function f(x)=x.

Starting at F(x). If you are given the point (-2,1), than you must have point (2,-1).
Do you have that point?

I hope I helped out.

Note: I am pretty confident I answered it correctly, if there are mistakes I hope the next guy corrects me, so I can pick up from there.
 
Last edited:
"one to one" means no two different inputs give the same output. since the outputs are apparently listed in the rows, "not one to one" means the row which has repeated entries. that would be the second row, for G.
 
JasonRox said:
Also, a function that is symmetrical about the origin is an odd function. An odd function is when f(-x)=-f(x).

I agree on your definition of odd function, but I would have said that even functions (such as G in the example above) are the ones best called 'symmetrical.' Anybody?
 
Yes, I agree with you Janitor.

To be even more specific one-to-one functions don't have an inverse function, or atleast at the level I am at right now.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top