Composite Functions, please confirm

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Homework Help Overview

The discussion revolves around composite functions, specifically the simplification of expressions involving the functions f(x)=4x-1, g(x)=3-2x^2, and h(x)=√(x+5). Participants express uncertainty regarding their calculations and seek confirmation on their work, particularly on the multiplication and simplification steps.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants share their attempts to simplify composite functions and express confusion over specific calculations, particularly in the context of squaring terms and handling square roots. Questions about the domain and range of the functions are also raised.

Discussion Status

Some participants provide feedback on the original poster's calculations, while others question the clarity of the lesson material. There is ongoing dialogue about the proper approach to determining the domain of the functions, with no clear consensus reached yet.

Contextual Notes

There are mentions of a homework template that should be used when posting questions, indicating that adherence to forum rules is part of the discussion. Additionally, some participants express frustration with the teaching quality of their math class, which may affect their understanding of the material.

dylanjames
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New to composite functions here. Lesson has been vague and unhelpful.. again. Here is what I've worked on so far but not sure on the last equation in particular, or that I have done my multiplication properly when working with a squared set of brackets, multiplied by an number.. (b and c)
Any confirmation/help would be mucho appreciated. Thanks!

Given the functions f(x)=4x-1, g(x)=3-2^2 and h(x)=√(x+5) simplify the following..

a) f(g(a))
g(a)=4(3-2a^2) - 1
=12 - 8a^2 - 1
=11-8a^2.

b) g(f(2x))
f(2x)=4(2x)-1=8x-1
g(f(2x)=3-2(8x-1)^2
g(f(2x)=3-2[(8x-1)(8x-1)]
g(f(2x))=3-2[64x^2 -16x+1]
g(f(2x)=-128x^2+32x-2+3
g(f(2x)=-128x^2+32x+1

c) h(g(2k+1))
g(2k+1)=3-2(2k+1)^2
=3-2[(2k+1)(2k+1)]
=3-2[4k^2 +4k+1]
=-8k^2-8k+1
h(g(2k+1))=sqrt[(-8k^2-8k+1)+5] =?
=-8k-2.82k+2.23? = 10.82k+2.23...
That can't be right...

Also need the domain and range for f(g(x))... no idea how to tackle that.
 
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dylanjames said:
New to composite functions here. Lesson has been vague and unhelpful.. again. Here is what I've worked on so far but not sure on the last equation in particular, or that I have done my multiplication properly when working with a squared set of brackets, multiplied by an number.. (b and c)
Any confirmation/help would be mucho appreciated. Thanks!

Given the functions f(x)=4x-1, g(x)=3-2x^2 and h(x)=√(x+5) simplify the following..
There's a typo in g(x).

a) f(g(a))
g(a)=4(3-2a^2) - 1
=12 - 8a^2 - 1
=11-8a^2.

b) g(f(2x))
f(2x)=4(2x)-1=8x-1
g(f(2x)=3-2(8x-1)^2
g(f(2x)=3-2[(8x-1)(8x-1)]
g(f(2x))=3-2[64x^2 -16x+1]
g(f(2x)=-128x^2+32x-2+3
g(f(2x)=-128x^2+32x+1

c) h(g(2k+1))
g(2k+1)=3-2(2k+1)^2
=3-2[(2k+1)(2k+1)]
=3-2[4k^2 +4k+1]
=-8k^2-8k+1
h(g(2k+1))=sqrt[(-8k^2-8k+1)+5] =?
=-8k-2.82k+2.23? = 10.82k+2.23...
That can't be right...

Also need the domain and range for f(g(x))... no idea how to tackle that.
##\displaystyle\ h(g(2k+1))=\sqrt{-8k^2-8k+1+5\ } ## is correct.

You can't split up the square root like that.
##\displaystyle\ \sqrt{a+b}\ne\sqrt{a}+\sqrt{b} ##​
 
So how do I solve the domain of that function?
 
dylanjames said:
So how do I solve the domain of that function?
What is meant by the implicit domain of a function, or meant by determining the domain of a function implicitly?
 
English dude!
 
dylanjames said:
English dude!
What sort of math class is this?

What is meant by the domain of a function anyway ?
 
An extremely poorly taught advanced functions class, that throws its students two small examples, slams them with 35 homework questions that go 6 steps further than taught and do me no good in terms of actually understanding material. I have solved the question. Thank you.
 
SammyS said:
What is meant by the implicit domain of a function, or meant by determining the domain of a function implicitly?

dylanjames said:
English dude!
What Sammy wrote was English. Chill on the attitude, dude, and show some respect, or else...
 
  • #10
Relax mark. Little bit of sarcasm got your knickers in a twist?
 
  • #11
dylanjames said:
Relax mark. Little bit of sarcasm got your knickers in a twist?
Sarcasm often doesn't come off well in a social media environment. In post #5 Sammy tried to get you to think about your question, and you replied with a flippant "English dude!" Whether you realize it or not, he was trying to help you. When he asked what type of class you were in, in post 7, I don't believe he was primarily interested in your opinion of how the class was run, but instead what the level of the class was.

You're the one asking for help. Sammy and others are helping completely voluntarily, so show some respect for those who are trying to help you understand this stuff. The "or else" means that this is your warning. There won't be others.
 
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  • #12
Lighten up. No disrespect to Sammy intended but I do believe this is why people have such a terrible perception of math-oriented people, and professors too for that matter.
Life is too short to sweat the little things Mark and ill be damned if I am going to sit on here and be lectured by some retired math professor, eager to relive his years as the "the man in charge" to compensate for his little man complex. Does it matter in the grand scheme of things whether each of the homework threads are posted in your perfect little columns? No. Does it make any difference that you send out lord knows how many emails to people complaining about their accidentally misplaced forums?? No.
Furthermore, does it matter that you are going to ban me from a forum that I can join in again under a new name in approximately 12 seconds?!?

Lighten up man. When the dust settles, you know I'm right. Cheers.
 
  • #13
dylanjames said:
Lighten up. No disrespect to Sammy intended but I do believe this is why people have such a terrible perception of math-oriented people, and professors too for that matter.
Life is too short to sweat the little things Mark and ill be damned if I am going to sit on here and be lectured by some retired math professor, eager to relive his years as the "the man in charge" to compensate for his little man complex. Does it matter in the grand scheme of things whether each of the homework threads are posted in your perfect little columns?
If you post a homework question, you need to use the template. Them's the rules. See https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/, second bullet under Homework Guidelines.
Attempts are mandatory
You MUST show that you have attempted to answer your question in order to receive help. You MUST make use of the homework template, which automatically appears when a new topic is created in the homework help forums.

You agreed to abide by these rules when you opened an account here. If you choose not to abide by them, your stay here will be short.
dylanjames said:
No. Does it make any difference that you send out lord knows how many emails to people complaining about their accidentally misplaced forums?? No.
Furthermore, does it matter that you are going to ban me from a forum that I can join in again under a new name in approximately 12 seconds?!?
You must think that no one has ever thought of this before. Just to set you straight, there have been a lot of members who were banned, and who signed up again with new user names. They didn't last long either.
dylanjames said:
Lighten up man. When the dust settles, you know I'm right. Cheers.
 
  • #14
dylanjames said:
Furthermore, does it matter that you are going to ban me from a forum that I can join in again under a new name in approximately 12 seconds?!?
It would not be beneficial to you, even if you sign up as a new user. You will get good responses to your questions only when you change to a good attitude and give respect.

You must know that some privileged members of PF can locate you by your IP address.
So changing up to a new user will not help you.
 
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