Doug C. answered • 10/30/21
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Doug C.
Is the video your professor posted available publicly, i.e. can I see it? If so, I will be willing to take a look to see if I am able to make sense of it and possibly clarify for you.
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10/30/21
Harbin S.
Yes sir, heres the private link, by just clicking here you will get access to view it. For the second equation i dont have a problem, i somehow figured it out how to solve it thanks to your effort. I just dont know whats the proper way of solving the first one, my mind is a mess with all these youtube videos solving it differently. You can clearly see that our professor is not capable of explaining problems in a proper way, now i dont mean it in a bad way but her english is bad and we get stuck throughout it, we find such it hard understanding the concepts of an exercise. Heres the link: https://www.youtube.com/watch?v=MdvLwU2aBVk ; Tell me what you think about it. Thank you again for responding to me, have a great rest of the day.
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10/30/21
Harbin S.
Yes sir, heres the private link, by just clicking here you will get access to view it. For the second equation i dont have a problem, i somehow figured it out how to solve it thanks to your effort. I just dont know whats the proper way of solving the first one, my mind is a mess with all these youtube videos solving it differently. You can clearly see that our professor is not capable of explaining problems in a proper way, now i dont mean it in a bad way but her english is bad and we get stuck throughout it, we find such it hard understanding the concepts of an exercise. Heres the link: https://www.youtube. com/watch?v=MdvLwU2aBVk (theres is a blank space on .com part, stick it and fix the link because i couldnt post the comment, it wouldnt let me share any kind of link)... Tell me what you think about it. Thank you again for responding to me, have a great rest of the day.
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10/31/21
Harbin S.
Where can i pass you the link? Can you give me your email or something because im trying to put it here on the comment but it wont actually let me, the website does not allow links or advertisers. Send me a platform or any social media that i can pass you the youtube link so you can check it. Thank you. Hopefully ill hear from you.
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10/31/21
Doug C.
You might be able to add the link if you leave off the http:// and even the www. Otherwise send it in pieces and I will put in the periods. I cannot send my personal email here.
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11/01/21
Harbin S.
Yes sir, heres the private link, by just clicking here you will get access to view it. For the second equation i dont have a problem, i somehow figured it out how to solve it thanks to your effort. I just dont know whats the proper way of solving the first one, my mind is a mess with all these youtube videos solving it differently. You can clearly see that our professor is not capable of explaining problems in a proper way, now i dont mean it in a bad way but her english is bad and we get stuck throughout it, we find such it hard understanding the concepts of an exercise. Heres the link: youtube/watch?v=MdvLwU2aBVk ; add the www and . com, tell me what you think about ti
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11/01/21
Doug C.
OK, I was able to view the video--which only goes over part b). Her strategy was only slightly different than mine in two ways. I chose to think of the range of f as the domain of its inverse, so I interchanged x and y before solving for y in terms of x. Once the roots were identified I chose to use the number line approach for fining when 7x^2 -18x +7 less then or equal to 0. She referenced its parabola and identified the interval where the parabola was below the x-axis. Really all the same.
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11/01/21
Doug C.
The video that I generated also addressed problem a). I interchanged x and y and then solved for y in terms of x to find the domain of the inverse function. This resulted in y = plus or minus the square root of (1-x) divided by square root (x). Watch my video again for part a) and let me know where it starts to get confusing. It seems like you might be having issues identifying the domain of a function based on the values of x that are not permitted. For example for 1/sqareroot(x), the domain must be x greater than 0 (because you can not take the square root of a negative number AND you cannot divide by zero. For the square root of (1-x) in the numerator you simply cannot take the square root of a negative number so 1-x greater than or equal to 0 which becomes x less than or equal to 1.
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11/01/21
Harbin S.
First off, i just wanna thank you a lot for taking your time and solving the equations in a proper way, thats very honest of you, thanks a lot sir. Theres just one question. How come you solved it in a 1 minute by finding the inverse function when theres my professor taking a whole 10-20min to solve a simple equation. And he didnt even used inverse function, it had something to do with right hand side and left hand side, now thats what i dont understand, because i end up watching videos in youtube but none of them are related to the answer that professor has given, i cant manage to find the same answer, i just end up with a lot of different answers in the end. Frustrating, im stuck here in this simple equation while im supposed to do other exercises, lol. Anyway thank you a lot sir. I really do appreciate it.10/30/21