A community for students.
Here's the question you clicked on:
 0 viewing
leozap1
 one year ago
*Fan & medal* Please explain how I solve this. I have no idea how to do this, when I tried I ended up with 30x^2  6.
Let f(x) = 5x + 3 and g(x) = 6x  2. Find f * g and its domain.
A.30x^2 + 28x  6; all real numbers except x = 1/3
B. 15x^2  8x  12; all real numbers except x = 3/5
C. 30x^2 + 28x  6; all real numbers
D. 15x^2  8x  12; all real numbers
leozap1
 one year ago
*Fan & medal* Please explain how I solve this. I have no idea how to do this, when I tried I ended up with 30x^2  6. Let f(x) = 5x + 3 and g(x) = 6x  2. Find f * g and its domain. A.30x^2 + 28x  6; all real numbers except x = 1/3 B. 15x^2  8x  12; all real numbers except x = 3/5 C. 30x^2 + 28x  6; all real numbers D. 15x^2  8x  12; all real numbers

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0probs a good idea to multiply the functions

leozap1
 one year ago
Best ResponseYou've already chosen the best response.0That's what I did, and I got 30x^2  6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f*g=(5x+3)(6x2)\] \[f*g=5x(6x2)+3(6x2)\] \[f*g=30x^2+10x+18x6\] \[f*g=30x^2+28x6\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we need to find the domain.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we can use to methods to find the domain; graphical method or algebra

leozap1
 one year ago
Best ResponseYou've already chosen the best response.0Yea I see what I did wrong thank you. I don't know how to find the domain though.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well here there are only two functions with our proposed solution right; answers A and C

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A says all real numbers except x=1/3 and C says x has solutions for all real numbers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so we say to our selves, if we plug x=1/3 into the equation, would we get a value for the function?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sure we do. f(x)=0 for x=1/3. but thats still a solution of the function right? that number exists. hence our answer is C

leozap1
 one year ago
Best ResponseYou've already chosen the best response.0Thanks! This really helps me. I also have some questions similar to this, but with division. Could you help me with it? Here is one of the questions. Let f(x) = x^2  16 and g(x) = x + 4. Find f/g and its domain. (1 point) A. x + 4; all real numbers except x =/ 4 B. x + 4; all real numbers except x =/ 4 C. x  4; all real numbers except x =/ 4 D. x  4; all real numbers except x =/ 4

leozap1
 one year ago
Best ResponseYou've already chosen the best response.0The =/ is suppose to be a does not equal to sign.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f/g=\frac{ x^216 }{ x+4 }\] \[f/g=\frac{ (x4)(x+4) }{ x+4 }\] \[f/g=x4\] and since the denominator of the previous step has x4, then a function does not exist at a certain point when the denominator equals zero. therefore \[f/g=x4\] \[\in R, \neq4\]

leozap1
 one year ago
Best ResponseYou've already chosen the best response.0This one's a lot easier than the first, thank you.

leozap1
 one year ago
Best ResponseYou've already chosen the best response.0Could you check my work on this last one? Let f(x) = 5x  4 and g(x) = 6x  7. Find f(x) + g(x). I got x11 for this one.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.