A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
I dont know how to find the inverse of this! Someone help me get it started?? Will give medals:
If g(x) = x^2 + 2x with x ≥ −1, find g^−1(15)=.
anonymous
 one year ago
I dont know how to find the inverse of this! Someone help me get it started?? Will give medals: If g(x) = x^2 + 2x with x ≥ −1, find g^−1(15)=.

This Question is Closed

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0find x in terms of g(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i just dont know how to start it, we never learned hwoo to do inverse functions with restrictions

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0Here there's a restriction because, without the restriction \(g^{1}(x)\) won't be a function...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know up to this point dw:1442069050216:dw

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0you're in the right path..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it wouldnt be a one to one function, I understand that part. i just dont know how to fnd this inverse. do i factor x^2+2x or...?

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0do you know how to solve quadratic equations?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442069191193:dw

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0not quite... Can you solve \(x^2+2xy=0\) for x?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im not sure how to do that one

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0does i matter that there is a y in the equation? and do i go about doing it with a=1, b=2 and c=1?

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0you've to treat y as a constant..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so a=1, b=2, c=1; is tat why the restriction says that x>=1? cause the square root of it wouldve made the radical a negative; and you cant take the square root of negative number (its imaginary)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442069827280:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait i messed that up: it should be

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442069945844:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442070088984:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442070113092:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442070133261:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so now what do i do, and is that correct? ^

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i dont think this is right.. can you show some work? that doesnt make sense

phi
 one year ago
Best ResponseYou've already chosen the best response.1use the quadratic formula, with a=1, b=2, c= y

phi
 one year ago
Best ResponseYou've already chosen the best response.1you should get \[ x= \frac{2\pm \sqrt{4+4y}}{2} \] which simplifies to \[ x= 1 \pm \sqrt{1+y} \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1now use the constraint that x \( \ge\) 1 that means we can't use the  sqr, otherwise x will be less than 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so that is the inverse of the function?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay t hat makes sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and so now I just plug in the 15 in for x?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no wait, dont i have to solve for y first or something?

phi
 one year ago
Best ResponseYou've already chosen the best response.1so now you have a function (one value for every y) \[ x= 1 + \sqrt{1+y}\] we now "switch x and y" because we assume x is the "input" \[ y = 1 + \sqrt{1+x}\]

phi
 one year ago
Best ResponseYou've already chosen the best response.1now you have \[ g^{1}(x) = 1 + \sqrt{1+x} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much, that makes alot of sense

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my professor never showed us how to do these kinds _ totally makes sense though, thanks again!

phi
 one year ago
Best ResponseYou've already chosen the best response.1the first trick was realizing you use the quadratic formula and treat y as the "c"

phi
 one year ago
Best ResponseYou've already chosen the best response.1then the second idea is to use x>=1 to know we don't use the sqrt as the solution i.e. we use 1 + sqrt()

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah thats what threw me off because weve been doing pproblems where we just have to solve for 'y', but we didnt do a different approach to solving it, i didnt know we could do that with by putting a variable in the quad formula. that definitely threw me through a loophole lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and we cant use 1 because that would make a nonreal number or imaginary number

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@phi quick question, how did you simplify under the radical??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what did you do in these 2 steps?dw:1442072831198:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442072887637:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.14+4y= 4(1+y) \[ \sqrt{4(1+y) }= \sqrt{4} \sqrt{1+y} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh wow, okay, thanks!

phi
 one year ago
Best ResponseYou've already chosen the best response.1btw \[ \sqrt{4+4y} \ne \sqrt{4}+ \sqrt{4y} \] for example, if y=1 we would have \[\sqrt{8} \ne 2+2 \\ \sqrt{8} \ne 4 \] as a calculator will show

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so those 2 are completely different, is it because its not held by multiplication?

phi
 one year ago
Best ResponseYou've already chosen the best response.1you can't split the sum inside a square root. you are allowed to factor into terms that multiply
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.