SFNQuestions Posted January 29, 2017 Share Posted January 29, 2017 Suppose that for solving for the inverse of a function that f(x) can be manipulated into the form of [math]g(f(h(x)))=x[/math]. Then, h(x) is inverted to show [math]g(f(x))=h^{-1}(x)[/math]. Afterwards, the function is reverted to show [math]h(g(f(x)))=x[/math]. Does this correctly show that h(g(x)) is in fact the inverse of f(x)? If not, what can be done in this situation to find the inverse of the original f(x)? Link to comment Share on other sites More sharing options...

wtf Posted January 29, 2017 Share Posted January 29, 2017 Can you give an example? Link to comment Share on other sites More sharing options...

SFNQuestions Posted January 29, 2017 Author Share Posted January 29, 2017 Can you give an example? That's okay, I have done some more testing and found this method is valid. But, thank you for inquiring, it could have just have easily been the case that this was too complicated for me to determine. Link to comment Share on other sites More sharing options...

wtf Posted January 29, 2017 Share Posted January 29, 2017 (edited) Well I'm glad I could help. Edited January 29, 2017 by wtf Link to comment Share on other sites More sharing options...

Xerxes Posted January 29, 2017 Share Posted January 29, 2017 Suppose that for solving for the inverse of a function that f(x) can be manipulated into the form of [math]g(f(h(x)))=x[/math]. Then, h(x) is inverted to show [math]g(f(x))=h^{-1}(x)[/math]. Afterwards, the function is reverted to show [math]h(g(f(x)))=x[/math]. Does this correctly show that h(g(x)) is in fact the inverse of f(x)? No, it is not valid (or only trivially). First some terminology..... The domain of a function is the set of all those elements that the function acts upon. Each element in the set is called an argument for the function The codomain - or range - of a function is the set of all elements that are the "output" of the function. So for any particular argument the element in the codomain is called the image of the argument under the function. So, Rule 1 for function...No element in the domain may have multiple images in the codomain. Rule 2 for functions.....Functions are composed Right-to-Left Rule 3 for functions.......Functions can be composed if and only if the codomain of a function is the domain of the function that follows it (i.e. as written, is "on the Left) Rule 4 for functions....... For some image in the codomain, the pre-image set is all those elements in the domain that "generate" this image. Notice that, although images are always single element, pre-images are sets - although they may be sets with a single member in which case the function is said to have an inverse - not otherwise. Look closely at what you wrote above. and check how many of these rules are violated. 2 Link to comment Share on other sites More sharing options...

studiot Posted January 29, 2017 Share Posted January 29, 2017 Well said Xerxes. +1 SFNrtc, I recommend drawing diagrams to help understand Xerxes comments. Link to comment Share on other sites More sharing options...

Country Boy Posted February 17, 2017 Share Posted February 17, 2017 Suppose that for solving for the inverse of a function that f(x) can be manipulated into the form of [math]g(f(h(x)))=x[/math]. Then, h(x) is inverted to show [math]g(f(x))=h^{-1}(x)[/math]. Afterwards, the function is reverted to show [math]h(g(f(x)))=x[/math]. Does this correctly show that h(g(x)) is in fact the inverse of f(x)? If not, what can be done in this situation to find the inverse of the original f(x)? No, it doe not show that h(g) is the inverse of f. You actually have to show two things- that h(g(f(x))= x and that f(h(g(x))= x. Link to comment Share on other sites More sharing options...

SFNQuestions Posted April 17, 2017 Author Share Posted April 17, 2017 (edited) No, it is not valid (or only trivially). First some terminology..... The domain of a function is the set of all those elements that the function acts upon. Each element in the set is called an argument for the function The codomain - or range - of a function is the set of all elements that are the "output" of the function. So for any particular argument the element in the codomain is called the image of the argument under the function. So, Rule 1 for function...No element in the domain may have multiple images in the codomain. Rule 2 for functions.....Functions are composed Right-to-Left Rule 3 for functions.......Functions can be composed if and only if the codomain of a function is the domain of the function that follows it (i.e. as written, is "on the Left) Rule 4 for functions....... For some image in the codomain, the pre-image set is all those elements in the domain that "generate" this image. Notice that, although images are always single element, pre-images are sets - although they may be sets with a single member in which case the function is said to have an inverse - not otherwise. Look closely at what you wrote above. and check how many of these rules are violated. Yeah I've found you're not entirely correct. You're obviously correct that not every function is invertable on every domain, but as a general technique it absolutely works because it's the same as doing a u-substitution and then back substituting it to find the inverse. I've done it tons of times now and it works every single time, long as I limit the domain. Edited April 17, 2017 by SFNQuestions Link to comment Share on other sites More sharing options...

Xerxes Posted April 17, 2017 Share Posted April 17, 2017 Yeah I've found you're not entirely correct.Then if you know better than those who try to guide you, you do not need to ask the question, right? Just a reminder: the image of your first function is [math]h(x)[/math]. Then you claim it's preimage is [math]h^{-1}(x)[/math]. So [math]h(x)=x[/math]. Agreed? And so on....... Link to comment Share on other sites More sharing options...

SFNQuestions Posted April 17, 2017 Author Share Posted April 17, 2017 Then if you know better than those who try to guide you, you do not need to ask the question, right? Just a reminder: the image of your first function is [math]h(x)[/math]. Then you claim it's preimage is [math]h^{-1}(x)[/math]. So [math]h(x)=x[/math]. Agreed? And so on....... Why? Because the foundation of math and science isn't in automatically assuming every single thing you think of is automatically correct. I'm not just going to make a statement and assume it's going to work out, I'm going to test a lot of times, I only asked it here in case I didn't have luck finding out the right answer or to see if anyone else could confirm the results. Link to comment Share on other sites More sharing options...

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