How to show that a Function is One-to-One algebraically | SHS 1 ELECTIVE MATH

This video discusses how to prove whether a function is one-to-one. A one to one function is the one where if the elements in the domain have distinct values in the codomain or range. For a function to be one to one, f(a) = f(b) such that a = b.
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#onetoonefunction
#domainofafunction
#rangeofafunction
Content
1. How to show whether a function is one to one or not.
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1 hour before exam moment

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[@SkanCityAcademy_SirJohn]

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[@SkanCityAcademy_SirJohn]

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[@benkiromat626]

I need a bit of clarification. For the first example, shouldn't the restriction be X not equal to -2 .

[@SkanCityAcademy_SirJohn]

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[@SkanCityAcademy_SirJohn]

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[@fatmahassan4402]

Is the function f(x)=x/y from z to z : one to one, onto or both ?
I do believe that it is not even a function , but the choices for this question in my exam didn’t even include this option.

[@zynex9250]

Thank you

[@SkanCityAcademy_SirJohn]

Most welcome

[@goodsonmumbelunga2331]

At 5:01 why didn't you also cancel the b² for it to remain b also?

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[@SkanCityAcademy_SirJohn]

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[@Urbanstrangler]

this is an oddly good explanation.

[@SkanCityAcademy_SirJohn]

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[@gourav1163]

4:50 why not also cancel sqrt of b with square that would give a = b hence one to one?

[@SkanCityAcademy_SirJohn]

you want to find the value of a, and a is equal to square root of b, square root of b, means there are two possible values of b. therefore a = -b or a = b

[@gourav1163]

@@SkanCityAcademy_SirJohn but won't I still get the value of a even if I cancel the sqrt with square

[@SkanCityAcademy_SirJohn]

Really?

[@SkanCityAcademy_SirJohn]

@@gourav1163 the square root of a value gives to possible values, eg sqr(4) gives two possible values, 2 and -2, 2*2 = 4, -2*-2 = 4.

[@felixorji385]

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[@SkanCityAcademy_SirJohn]

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[@Birrby]

5:17 why can’t we cancel both square roots so they’re both equal to +-A and +-B why cancel A only is there anything specific?

[@SkanCityAcademy_SirJohn]

No please, we can't do that. We can only cancel inside which is usually b. So as to find a.

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[@SkanCityAcademy_SirJohn]

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[@SkanCityAcademy_SirJohn]

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[@arnobbhattacharjee1853]

Thank you so much. But I have a question. Why didn't you take + - a just like you did for root b ?

[@SkanCityAcademy_SirJohn]

you just need to make one variable the subject, and that's going to be positive. For example if you have a² = sqr(4), then you say a = +/-2.

[@arnobbhattacharjee1853]

@@SkanCityAcademy_SirJohn Thank you for replying. One more question sir. At 6:07 it was said that x > 0. That's why you didn't consider a = - b. But if it was said that domain of the function can both be negative and positive, would the function be one to one then?

[@SkanCityAcademy_SirJohn]

Yes, it would have been one to one

[@arnobbhattacharjee1853]

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thnk u ,now how can i prove this proposition Let f : A −→ B be a mapping. We have f ◦ 1A = 1B ◦ f.(how can i prove this)

[@SkanCityAcademy_SirJohn]

Please can you make your question more clearer, so I can help you out?

[@emoney752]

how is a+-b not 1-1 in the second example but it is in the 3rd example when in both cases they are a+-b?

[@SkanCityAcademy_SirJohn]

This is because in question c, a condition is specified that x > 0, which means a cannot be a negative answer. Hence a = -b cannot hold. But for question b, there is no constraints on x. So a can be any number. So once a = -b at some point it means, the f is not one to one

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[@hamahawre3635]

For the second example, wht we can't cancel square root 2 with the radical (b)??

[@SkanCityAcademy_SirJohn]

Thanks so much for your question, but i dont really get it. kindly state the exact time in the video, so i check it out for you.

[@hamahawre3635]

Thanks I got it, but what should we do if the function is not a fraction??

[@SkanCityAcademy_SirJohn]

@hamahawre3635 you just substitute a on one side and b on the other side

[@vimalkumar-iw9du]

Show that mapping f:R->R defined by f(x)=3x+5 is one one into where R is the set of real numbers.
Can you please tell me solution for checking into function on this linear equation x=(y-5)/3

[@SkanCityAcademy_SirJohn]

Can you please come again with your question?

[@vimalkumar-iw9du]

Show that mapping f:R->R defined by f(x)=3x+5 is one one into where R is the set of real numbers

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[@zying9986]

May I ask if the answer is -a=-b then will it be a one to one function?

[@SkanCityAcademy_SirJohn]

yes, it is one to one function, because you can divide through by negative 1, then finally a will be equal to b, a = b

[@SkanCityAcademy_SirJohn]

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[@heartbeldad3005]

Why is that the root of a is not +_ a just like b? How can you know the root of variables?

[@SkanCityAcademy_SirJohn]

Well, for a function to be one to one, the condition f(a) = f(b), a=b must be satisfied, so the attempt is to make a the subject, so a = √b². now b² can either be (b*b) or (-b*-b) which means that a=b, or a=-b. So since the condition a = b is not satisfied at all times for the function, it is not one to one.

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[@noisemelon]

wouldnt a always equal b even if the function wasn't 1 to 1 because the equation equals the same equation?

[@SkanCityAcademy_SirJohn]

can you specify the time stamp for your question in the video?

[@Random_guy909]

I may not be 100% correct. But I'll still write what i understood.
to be a one to one function you need one x assigned to one unique y or f(x)
only one-to-one functions can have inverse functions.
[if a function has (x,y) point, the inverse function would return (y,x). Basically fliping the inputs and outputs]
Let's take this example: F(x) = 1/(x+2)
Domain. Range
A B
1 1/3
2 1/4
3 1/5
a 1/(a+2)
b 1/(b+2)
If f(a) and the f(b) are the same then then 1/(a+2) and 1/(b+2) are the same. Thus in this example the input would be (a=b). if we were to put (a, 1/(a+2)) point the inverse would be (1/(a+2) , a)
The same goes for ( b, 1/(b+2)) and it's inverse.
2nd example:
{ ( a, f(a)) , ( b, f(b))}
Where, a doesn't equal to b
And f(a) = f (b)
in any example if a does not equal to b then
Domain. Range
a
f(a) or f(b)
b [since f(a) = f(b)]
Because a and b are different, they are different input but f(a) and f(b) are equal. So 2 points of this function would be ( a, f(a)) and ( b, f(a)). The inputs only have one output so this is a function. But when we try to inverse, it'd be
( f(a) , a) and ( f(a) , b)
Here we can see that the inputs are the same but outputs are different. Thus the inverse is not a function.
In conclusion, the function we mainly began with doesn't have an inverse function.
Since, only one to one function can have inverse function. Therefore the example 2 is not a one-to-one function.

[@SkanCityAcademy_SirJohn]

@@Random_guy909 yes you are correct, i think i got that in the video right, a = b, a = -b, hence not one to one?

[@madsartsy]

I am a bit confused- how does the square root of a^2 = a, while the square root of b^2 = +/- b?
7:20

[@SkanCityAcademy_SirJohn]

You shouldn't be confused you want to find a, so a will definitely be equal to plus or minus b.

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Oh okay. You are welcome

[@akwasiamoako1952]

4:57 ...why didn't the square cancel the square root of b just as you did in a.....so that it becomes only a=b

[@SkanCityAcademy_SirJohn]

That can't be because, you want to find a , and
a = √b²
Thus, we have two, both the positive and negative values of b.
Since we want to find a ,we cant make a plus or minus. But rather b

[@akwasiamoako1952]

@@SkanCityAcademy_SirJohn thank u sir..... it's clear

[@flexcheekz4582]

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[@SkanCityAcademy_SirJohn]

@flexcheekz4582 that's great

[@thamashadevindi3564]

F(x)= square root of x+3x , x>=-3 can you explain how to solve this

[@SkanCityAcademy_SirJohn]

hi Thamasha, please come again with your question, are you sure its f(x) = x + 3x??

[@thamashadevindi3564]

@@SkanCityAcademy_SirJohn it's square root of (x+3x)

[@thamashadevindi3564]

I want to know how to find this function is one-to-one function or no

[@SkanCityAcademy_SirJohn]

@@thamashadevindi3564 f(x) = square root of x+3x, f(a) = f(b), square root of a + 3a = square root of b + 3b, square both sides, 4a = 4b, a = b, since f(a) = f(b), the function is one-to-one

[@SkanCityAcademy_SirJohn]

@@thamashadevindi3564 it is one to one

[@hafizsahran791]

why f(a)= f(b) ... i'm so fool dont get it

[@SkanCityAcademy_SirJohn]

that is the condition for a function that is one to one, can you tell me the question number that you have an issue with?

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