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x3_drummerchick
 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)=.
x3_drummerchick
 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)=.

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

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1i 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...

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1I 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..

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1it 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?

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1dw: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?

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1im not sure how to do that one

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1quadratic formula?

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1does 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..

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1so 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)

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442069827280:dw

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1wait i messed that up: it should be

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442069945844:dw

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442070088984:dw

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442070113092:dw

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442070133261:dw

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1so now what do i do, and is that correct? ^

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1@Jhannybean help :(

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1i 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

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1so that is the inverse of the function?

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1oh okay t hat makes sense

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1and so now I just plug in the 15 in for x?

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1no 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 \[ g^{1}(x) = 1 + \sqrt{1+x} \]

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1thank you so much, that makes alot of sense

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1my 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()

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1yeah 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

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1and we cant use 1 because that would make a nonreal number or imaginary number

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1@phi quick question, how did you simplify under the radical??

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1what did you do in these 2 steps?dw:1442072831198:dw

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1dw:1442072887637:dw

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

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1oh 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

x3_drummerchick
 one year ago
Best ResponseYou've already chosen the best response.1so 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
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