I think you meant to get rid of the other side of the parabola which is contained within ]-∞ ; 3/4[ as there are a lot of ways to prove that [3/4 ; ∞[ is the right domain of restriction to be taken for f(x) to have that particular inverse. For example the inverse function always has the same monotony of the base function in the domain restriction taken. Pls correct me if I'm wrong
In the end my inverse is not a function; it is a multifunction. I do this to recover the entire parabolic shape of the original. I believe your thoughts are correct for the standard procedure to derive a true inverse function. This, like sine and cosine, requires a domain restriction.
Great question. If a function has an inverse, it is a reflection about y=x. However, many functions cannot be reflected and maintain their definition as a function, because the reflection fails the 'vertical line test'. To overcome some of these issues, typically a domain restriction is used to cut-down the function to something that will reflect and stay a function. Doing this will change the plots that can have an inverse for. If we throw out the requirement to be a function, so we can keep the original plots, then we need something like I use, which is a multifunction. The exercises that are typically seen in high school ask you to determine a domain restriction, and then find the inverse of this new function. I hope this is helpful. Cheerful Calculations. 🧮
I don't know of a direct route between g(x) and f'(x). However, since f(x) is given and we are told g(x), its inverse exists; we have the domain of f(x) is a subset of the Real numbers and is a bijective function. Furthermore, f(x) is continuous and increasing. We can proceed to differentiate f(x) determining f'(x) = 2+ (sqrt{2x-8})^{-1}.
f(x) = 2x²-3x-5 f⁻¹(x) = y f(f⁻¹(x)) = x f(y) = x 2y²-3y-5 = x 2y²-3y-(5+x) = 0 y = (3±√(9+4•2•(5+x))/4 We throw out the negative branch because we need f⁻¹(x) to be single-valued, do some algebra and get: f⁻¹(x) = 1/4•√(8x+49)+3/4
Have you studied left inverses and right inverses? Do me a favor and use your f(x) and f⁻¹(x) to compose f⁻¹(f(x)) keeping only the positive branch. What result did you get? Is this the expectation?
I don't know if that is easier? One of the great things about mathematics is that; there is usually more than one way to get to the correct solution. This allows people can take the avenue which works best for them. 🧮
I don't want to encourage you, or anyone, to consider themselves dumb. Mathematics is something that requires we spend time with to build our skills. We need time and practice to get better and find success. I'm glad this video was understandable. Have a great day!
Wouldn't it have been easier to complete the square on the original function to get it into the vertex form and THEN proceed with the inverse operation steps? Of course, you wind up in the same place, but most students know how to complete the square for a quadratic as a stand alone process. Also, when you put it into vertex form, it's easier to see the necessary domain restrictions because you have the vertex right in front of you. Very thorough video. Well done.
I don't know if that is easier? One of the great things about mathematics is that; there is usually more than one way to get to the correct solution. This allows people can take the avenue which works best for them.
Neither of your equations are functions which are one-to-one and onto (bijective). This means, like my example of a quadratic, there is no true inverse function. At best I could find a multifunction that would act as the inverse of the onto function I used as my example, the quadratic. Your first example might have a similar work-around. The second example you have chosen is discrete, mine was continuous. These are interesting problems to consider!
I think you meant to get rid of the other side of the parabola which is contained within ]-∞ ; 3/4[ as there are a lot of ways to prove that [3/4 ; ∞[ is the right domain of restriction to be taken for f(x) to have that particular inverse. For example the inverse function always has the same monotony of the base function in the domain restriction taken. Pls correct me if I'm wrong
In the end my inverse is not a function; it is a multifunction. I do this to recover the entire parabolic shape of the original. I believe your thoughts are correct for the standard procedure to derive a true inverse function. This, like sine and cosine, requires a domain restriction.
Why does the inverse not mirror the function on the plane?
Great question. If a function has an inverse, it is a reflection about y=x. However, many functions cannot be reflected and maintain their definition as a function, because the reflection fails the 'vertical line test'. To overcome some of these issues, typically a domain restriction is used to cut-down the function to something that will reflect and stay a function. Doing this will change the plots that can have an inverse for. If we throw out the requirement to be a function, so we can keep the original plots, then we need something like I use, which is a multifunction. The exercises that are typically seen in high school ask you to determine a domain restriction, and then find the inverse of this new function. I hope this is helpful. Cheerful Calculations. 🧮
@@TranquilSeaOfMath Thank you for your service
Ok say that we have f with an inverse function g
How can we prove that if f is rising, g is also rising?
Please find my response at ua-cam.com/video/-zpQT8DtHP0/v-deo.html
I look forward to your feedback.
If we have a function
f(x)=2x + sqr[2x-8], with X € [4, -> }
and we know it has an inverse g(x).
How can we find f’(x) through the inverse g(x)?
I don't know of a direct route between g(x) and f'(x). However, since f(x) is given and we are told g(x), its inverse exists; we have the domain of f(x) is a subset of the Real numbers and is a bijective function. Furthermore, f(x) is continuous and increasing. We can proceed to differentiate f(x) determining f'(x) = 2+ (sqrt{2x-8})^{-1}.
f(x) = 2x²-3x-5
f⁻¹(x) = y
f(f⁻¹(x)) = x
f(y) = x
2y²-3y-5 = x
2y²-3y-(5+x) = 0
y = (3±√(9+4•2•(5+x))/4
We throw out the negative branch because we need f⁻¹(x) to be single-valued, do some algebra and get:
f⁻¹(x) = 1/4•√(8x+49)+3/4
Have you studied left inverses and right inverses? Do me a favor and use your f(x) and f⁻¹(x) to compose f⁻¹(f(x)) keeping only the positive branch. What result did you get? Is this the expectation?
Wouldn't be easier to use completing the square and then getting the vertex then finding the inverse ?
I don't know if that is easier? One of the great things about mathematics is that; there is usually more than one way to get to the correct solution. This allows people can take the avenue which works best for them. 🧮
YOU understand how to talk to people like me, dumdums. You can speak my language. And I can understand simply what is being done. Thank you.
I don't want to encourage you, or anyone, to consider themselves dumb. Mathematics is something that requires we spend time with to build our skills. We need time and practice to get better and find success. I'm glad this video was understandable. Have a great day!
Wouldn't it have been easier to complete the square on the original function to get it into the vertex form and THEN proceed with the inverse operation steps? Of course, you wind up in the same place, but most students know how to complete the square for a quadratic as a stand alone process. Also, when you put it into vertex form, it's easier to see the necessary domain restrictions because you have the vertex right in front of you. Very thorough video. Well done.
I don't know if that is easier? One of the great things about mathematics is that; there is usually more than one way to get to the correct solution. This allows people can take the avenue which works best for them.
Hello sir, how about this?
y=3x³ - 5x² - 6x - 10
This too please
y²x + 4yx - 2x = 2y² - 4
Neither of your equations are functions which are one-to-one and onto (bijective). This means, like my example of a quadratic, there is no true inverse function. At best I could find a multifunction that would act as the inverse of the onto function I used as my example, the quadratic. Your first example might have a similar work-around. The second example you have chosen is discrete, mine was continuous. These are interesting problems to consider!
The start music is creepy
It's not meant to be creepy. It is my own music though.
Ok
Oh man I love it! What you are hearing is that it is out of tune, but somehow it sounds cool.
The intro is so out of tune but sounds cool. First time watcher...
Definitely out of tune. It is my own though.
@@TranquilSeaOfMath Unique. I like it!
Thanks. I like to do my own work and not pirate everything. I like to have things I create to call my own.
@@TranquilSeaOfMath Good video as well. I am a retired engineer and enjoying relearning math.
subscribed😀
Glad to see you found me!