x3_drummerchick
  • x3_drummerchick
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)=.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amilapsn
  • amilapsn
find x in terms of g(x)
x3_drummerchick
  • x3_drummerchick
i just dont know how to start it, we never learned hwoo to do inverse functions with restrictions
amilapsn
  • amilapsn
Here there's a restriction because, without the restriction \(g^{-1}(x)\) won't be a function...

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x3_drummerchick
  • x3_drummerchick
I know up to this point |dw:1442069050216:dw|
amilapsn
  • amilapsn
you're in the right path..
x3_drummerchick
  • x3_drummerchick
it 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
  • amilapsn
do you know how to solve quadratic equations?
x3_drummerchick
  • x3_drummerchick
yes
x3_drummerchick
  • x3_drummerchick
|dw:1442069191193:dw|
amilapsn
  • amilapsn
not quite... Can you solve \(x^2+2x-y=0\) for x?
x3_drummerchick
  • x3_drummerchick
im not sure how to do that one
x3_drummerchick
  • x3_drummerchick
quadratic formula?
amilapsn
  • amilapsn
yep
x3_drummerchick
  • x3_drummerchick
does 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
  • amilapsn
you've to treat y as a constant..
x3_drummerchick
  • x3_drummerchick
so 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
  • x3_drummerchick
|dw:1442069827280:dw|
x3_drummerchick
  • x3_drummerchick
wait i messed that up: it should be
x3_drummerchick
  • x3_drummerchick
|dw:1442069945844:dw|
x3_drummerchick
  • x3_drummerchick
|dw:1442070088984:dw|
x3_drummerchick
  • x3_drummerchick
|dw:1442070113092:dw|
x3_drummerchick
  • x3_drummerchick
|dw:1442070133261:dw|
x3_drummerchick
  • x3_drummerchick
so now what do i do, and is that correct? ^
x3_drummerchick
  • x3_drummerchick
@Jhannybean help :(
x3_drummerchick
  • x3_drummerchick
@amilapsn
amilapsn
  • amilapsn
c=-y
x3_drummerchick
  • x3_drummerchick
i dont think this is right.. can you show some work? that doesnt make sense
phi
  • phi
use the quadratic formula, with a=1, b=2, c= -y
phi
  • phi
you should get \[ x= \frac{-2\pm \sqrt{4+4y}}{2} \] which simplifies to \[ x= -1 \pm \sqrt{1+y} \]
phi
  • phi
now use the constraint that x \( \ge\) -1 that means we can't use the - sqr, otherwise x will be less than -1
x3_drummerchick
  • x3_drummerchick
so that is the inverse of the function?
x3_drummerchick
  • x3_drummerchick
oh okay t hat makes sense
x3_drummerchick
  • x3_drummerchick
and so now I just plug in the 15 in for x?
x3_drummerchick
  • x3_drummerchick
no wait, dont i have to solve for y first or something?
phi
  • phi
so 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" \[ y = -1 + \sqrt{1+x}\]
x3_drummerchick
  • x3_drummerchick
ohhhhh okay
phi
  • phi
now you have \[ g^{-1}(x) = -1 + \sqrt{1+x} \]
x3_drummerchick
  • x3_drummerchick
thank you so much, that makes alot of sense
x3_drummerchick
  • x3_drummerchick
my professor never showed us how to do these kinds -_- totally makes sense though, thanks again!
phi
  • phi
the first trick was realizing you use the quadratic formula and treat -y as the "c"
phi
  • phi
then 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
  • x3_drummerchick
yeah 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
  • x3_drummerchick
and we cant use -1 because that would make a nonreal number- or imaginary number
x3_drummerchick
  • x3_drummerchick
@phi quick question, how did you simplify under the radical??
x3_drummerchick
  • x3_drummerchick
what did you do in these 2 steps?|dw:1442072831198:dw|
x3_drummerchick
  • x3_drummerchick
|dw:1442072887637:dw|
phi
  • phi
4+4y= 4(1+y) \[ \sqrt{4(1+y) }= \sqrt{4} \sqrt{1+y} \]
x3_drummerchick
  • x3_drummerchick
oh wow, okay, thanks!
phi
  • phi
btw \[ \sqrt{4+4y} \ne \sqrt{4}+ \sqrt{4y} \] for example, if y=1 we would have \[\sqrt{8} \ne 2+2 \\ \sqrt{8} \ne 4 \] as a calculator will show
x3_drummerchick
  • x3_drummerchick
so those 2 are completely different, is it because its not held by multiplication?
phi
  • phi
you can't split the sum inside a square root. you are allowed to factor into terms that multiply
x3_drummerchick
  • x3_drummerchick
oh okay

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