What type of function is this ?

[alijan kk]

Homework Statement

MCQS is in the image,

Homework Equations

The Attempt at a Solution

as the range of this function is the subset of B , it is an injective function ?

option one is correct ?

 

[.Scott]

I believe it is injective but not surjective.
The domain is defined as -0.5 to +0.5.
The resulting range is -0.4 to +0.4.
So every value in the domain yields a unique value in the range (injective).
But there are some values in the potential range (-0.5 to -0.4 and 0.4 to 0.5) which are not used.

 

[DrClaude]

The domain is defined as -0.5 to +0.5.
The resulting range is -0.4 to +0.5.

Really?

@alijan kk: What are the domain and the range of the function? What is the meaning of injective and surjective?

 

[Dick]

I believe it is injective but not surjective.
The domain is defined as -0.5 to +0.5.
The resulting range is -0.4 to +0.5.
So every value in the domain yields a unique value in the range (injective).
But there are some values in the potential range (-0.5 to -0.4 and 0.4 to 0.5) which are not used.

Aside from the fact that is wrong, I would like to hear what alijan kk thinks the range is, and why.

 

[.Scott]

Really?

I fixed it.

 

[DrClaude]

I fixed it.

No, still wrong. But please let the OP answer now (this is the homework forum).

 

[alijan kk]

No, still wrong. But please let the OP answer now (this is the homework forum).

the B set is
(-0.5,0.5) f=b is the range
if the Ran f does not equal to B, then f is said to be into function , but this function is also one to one ,, so called injective ? thanks for the reply

 

[DrClaude]

but this function is also one to one ,, so called injective ?

But is it one-to-one? Do you get a unique value for each $x$ in the domain of $f$?

You should also check if it is surjective or not.

 

[alijan kk]

But is it one-to-one? Do you get a unique value for each $x$ in the domain of $f$?

You should also check if it is surjective or not.

it is not ijective because, the range is not a subset ! "serjective but not injective" should be correct ? and serjective has range=B [-0.5,0.5]

 

[Mark44]

MCQS is in the image

What is "MCQS"?

 

[Mark44]

But please let the OP answer now (this is the homework forum).

Noted...

 

[alijan kk]

Noted...

multiple choice question

 

[alijan kk]

Noted...

it is therefore requested kindly give me a hint or tell me the answer

 

[Mark44]

it is therefore requested kindly give me a hint or tell me the answer

Post #3 is a hint. A graph of this function would be helpful.

By forum rules, we can steer you in the right direction, but we can't just tell you the answer.

 

[alijan kk]

Post #3 is a hint. A graph of this function would be helpful.

By forum rules, we can steer you in the right direction, but we can't just tell you the answer.

domain is all real numbers,, and range is [-0.5,0.5]

 

[Mark44]

domain is all real numbers,, and range is [-0.5,0.5]

If you solve the equation $y = \frac 1 {1 + x^2}$ for x, do you get a single value of x for each y value? If so, the original function is invertible, and therefore one-to-one (i.e., injective).
If there is at least one x value for each number in the range, then the function is surjective.

 

[SammyS]

Homework Statement

MCQS is in the image,

Homework Equations

The Attempt at a Solution

as the range of this function is the subset of B , it is an injective function ?

option one is correct ?

Your thread would be improved if you were to display the image, rather than just showing the thumb-nail.

Like this:

 

[Physics345]

In your text there is a section for this clearly explaining how to find the proper range and domain and ******restrictions****** which is very important, I suggest you keep reviewing. Mastering your subject is key, that's when you really see the real benefits and understanding of Math.

 

[mathwonk]

an important tool for proving "surjectiveness" is the intermediate value theorem. do you have this available? it requires your function to be "continuous". on an interval domain. If so then it says that whenever you find two points of the target interval that are in the image of the function, then you can be sure that the whole interval between those two values is also in the image of the function. (In fact with a function like this one, there is really no other way to do this problem rigorously. I.e. even if you solve directly for the inverse images of potential values using root functions, you need that theorem to guarantee that the relevant roots actually exist. Of course you may be assuming that fact. I.e. do you know that square roots of non negative real numbers always exist? maybe they assume that in elementary courses.)

so either try a direct approach assuming roots exist, or try using that theorem, is my suggestion. in fact you might benefit from looking up that theorem just to expand your education, if it seems unknown. You might try as an exercise proving with that theorem that if f is a continuous function on R, and if points a,b,c exist with a<b<c, and with f(a) < f(b), and also f(c) < f(b), then f cannot be injective.i claim that if you use the intermediate value theorem, and look at the two values f(1) and f(-1), your problem is solved. it is up to you to see why.