Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

slaaibakBest ResponseYou've already chosen the best response.1
f . g = h f = 3x + 5 h = 3x^2 + 3x + 2 so f(g(x)) = 3x^2 + 3x + 2 That means: 3(g(x)) + 5 = 3x^2 + 3x + 2 By inspection: g(x) = x^2 + x  1
 one year ago

Studentc14Best ResponseYou've already chosen the best response.0
what's inspection, sorry?
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
Figuring it out by looking at it
 one year ago

Studentc14Best ResponseYou've already chosen the best response.0
I don't follow, how you get this part? 3(g(x)) + 5 = 3x^2 + 3x + 2
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
we know h(x) = 3x^2 + 3x +2 > (1) and we know f.g = f(g(x)) but f(x) = 3x + 5, therefore f(g(x)) = 3(g(x)) + 5 > (2) and we know that f.g = h > f(g(x)) = h(x) therefore: (1) = (2) 3(g(x)) + 5 = 3x^2 + 3x +2
 one year ago

Studentc14Best ResponseYou've already chosen the best response.0
ohkay, and then you can kinda just look at it and like guess and check and see that x^2+x1 works for g, right?
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
Yes. Or you can just solve G like a normal variable: 3(g(x)) + 5 = 3x^2 + 3x +2 Substract 5 on both sides: 3(g(x)) = 3x^2 + 3x 3 Divide by 3: g(x) = x^2 + x  1
 one year ago

Studentc14Best ResponseYou've already chosen the best response.0
oh! that's much clearer. thank you so much!
 one year ago

slaaibakBest ResponseYou've already chosen the best response.1
haha, colloquial for "cool, no problem"
 one year ago
See more questions >>>
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.