This is a very good question. It is not injective - test y =1 with the horizontal line test. To prove surjectivity we have to know the co-domain. Send a picture of the question to promathacademy1@gmail.com
By adding 10 and subtracting 10, this allows us to simply the expression by writing one as x divided by x. It is a manipulation technique used in algebra used to get the expression in a certain form.
adding 10 and subtracting 10 is a mathematical expression which keeps the previous expression the same. Here we needed a multiple of the denominator. (a-2) so we use 5(a-2), which is 5a-10. However we only have 5a+1 but we need 5a-10 somehow, so we subtract 10 and added 10 so that 5(a-2)/(a-2) simplifies to 5 and 5(b-2)/(b-2) simplifies to 5 which simplifies the whole equation.
for the second example, shouldn't |x| > 0 but not >= due to natural number? Im still very confused on this topic I hope you could clear my confusion.. thanksss
|x| >=0 for this question. Remember x comes from the domain and here the domain is Z which includes 0. The function is f(x) =|x|+2 so .........f(x) >0. |x| is a function but it is not the subject of the question by itself. Always consider what is your domain. Let me know that answers the question.
@@stevencheng7997 |x|+2>=2. Remember this depends on the domain of your function. I cannot emphasize this enough. For question 2 our domain was all integers but if the domain changed to "natural numbers" the result would be different.
I thought that when we have absolute value, the results will be always positive. So if we put -1 inside the absolute value, the results will be 1. So i got confuse in the question no. 4, i thought it was injective haha. Please correct me if i'm wrong😊
@@promathacademy8512 My bad. I should have been more specific as you said. I am referring to why you still had (a-5) or (5-a) on the left hand side if you had already divided it on the right hand side.
@@IKdeoSSa_7 yes we realize. Thank you for bringing that to our attention. The (a-5) on the left should have being canceled. We were rushing to get to the end. 😅
The range seem to be [3, ∞) while the codomain is [2, ∞) so since the range is not identical to the co domain you can conclude the the function is not surjective.
this really helped me a lot for my exam which is tmr , loved your explanation
Thank you so much, this tutorial is the most helpful of all the other videos I came across
That's just crazy....but thank you very much for letting us know, please consider subscribing.
You're the best! A REAL EXAMPLE
We love and appreciate your support!
Thanks this by far the clearest video , def subscribing
This was hella useful lmao
Have a test next week
Thank you so much
Thank you for your comment. It was our pleasure.
Great examples and excellent explanation. Thanks!
Glad you found it helpful. Please consider subscribing to help the channel grow. Every subscription counts and we appreciate it so much!
20:53 where did you get the "divided by a-5" from? x (5 - a) = -1 - 2 a, so x = (-1 - 2a) / (5-a)
i'd also like to know what the heck was done there and how that means it's not 5. Sir
This is what I was looking for.Thnx .Very well explained.
Thank you for you immeasurable support.
Good video👍
Proving that a linear transformation is bijective seens so much easier than proving that a normal function like y=x+3 is bijective for some reason
i do not understand why or how did you divide by a-5 at 20:50
Thank you got a better understanding of this concept 👍
I think by modulus you meant absolute value, but the overall content is understood, thank you.
That is correct.
Amazing video!!
thank you so much it was helpful
for the question 3 / the function is surjective since its already written that R - {2} this is read as " for all R numbers except number 2"
Yes, that is correct!
Love to see our subscribers interacting. Just to answer your question, your question is already answered ! Keep up the great work.
exellent video, very transparent
Very useful
in the last proof example why did u divide by a-5 instead of 5-a?
I asked the same question. He said it was an error.
ngl ur handwriting is beautiful
How would this work for functions with multiple expressions, such as f(x) = { x + 1 if x is even, x - 3 if x is odd } ?
This is a very good question. It is not injective - test y =1 with the horizontal line test. To prove surjectivity we have to know the co-domain. Send a picture of the question to promathacademy1@gmail.com
@@promathacademy8512 thanks, I've just emailed it to you
@@jazzysocksdude we sent the solution to your email. The function is bijective!!!! 😁
love you boss
Why did you add 10 and subtract 10 in the last question?
By adding 10 and subtracting 10, this allows us to simply the expression by writing one as x divided by x. It is a manipulation technique used in algebra used to get the expression in a certain form.
This video deserves millions of views 🔥🤍🤍🤍 , thank you so much
Where did you get your 10 from
At 17:07 where did the 10s come from?
adding 10 and subtracting 10 is a mathematical expression which keeps the previous expression the same. Here we needed a multiple of the denominator. (a-2) so we use 5(a-2), which is 5a-10. However we only have 5a+1 but we need 5a-10 somehow, so we subtract 10 and added 10 so that 5(a-2)/(a-2) simplifies to 5 and 5(b-2)/(b-2) simplifies to 5 which simplifies the whole equation.
for the second example, shouldn't |x| > 0 but not >= due to natural number? Im still very confused on this topic I hope you could clear my confusion.. thanksss
|x| >=0 for this question. Remember x comes from the domain and here the domain is Z which includes 0. The function is f(x) =|x|+2 so .........f(x) >0. |x| is a function but it is not the subject of the question by itself. Always consider what is your domain. Let me know that answers the question.
@@promathacademy8512 ohhhh snap yes |x| can be >= 0 but not |x|+2 >=, is that right
@@stevencheng7997 |x|+2>=2. Remember this depends on the domain of your function. I cannot emphasize this enough. For question 2 our domain was all integers but if the domain changed to "natural numbers" the result would be different.
I thought that when we have absolute value, the results will be always positive. So if we put -1 inside the absolute value, the results will be 1. So i got confuse in the question no. 4, i thought it was injective haha. Please correct me if i'm wrong😊
20:53 is my doubt
So a=5? Be specific ?
@@promathacademy8512 My bad. I should have been more specific as you said. I am referring to why you still had (a-5) or (5-a) on the left hand side if you had already divided it on the right hand side.
@@IKdeoSSa_7 yes we realize. Thank you for bringing that to our attention. The (a-5) on the left should have being canceled. We were rushing to get to the end. 😅
@@promathacademy8512 You are welcome👍
(i) Prove that the function f : [1, ∞) −→ [2, ∞) given by f(x) = (x + 2)
is not surjective.(please help)
The range of the function is not equal to the codomain. Simple ! Find the range of x+2
The range seem to be [3, ∞) while the codomain is [2, ∞) so since the range is not identical to the co domain you can conclude the the function is not surjective.
Very informative, but the sound of you sucking on a candy was really annoying.
didn't get anything
Were you eating candy while speaking? The sound is annoying to be honest.
Sorry to say that it is a very poor presentation 😢
Please control your mouth noises