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

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- amilapsn

find x in terms of g(x)

- x3_drummerchick

i just dont know how to start it, we never learned hwoo to do inverse functions with restrictions

- amilapsn

Here there's a restriction because, without the restriction \(g^{-1}(x)\) won't be a function...

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## More answers

- x3_drummerchick

I know up to this point
|dw:1442069050216:dw|

- amilapsn

you're in the right path..

- 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

do you know how to solve quadratic equations?

- x3_drummerchick

yes

- x3_drummerchick

|dw:1442069191193:dw|

- amilapsn

not quite... Can you solve \(x^2+2x-y=0\) for x?

- x3_drummerchick

im not sure how to do that one

- x3_drummerchick

quadratic formula?

- amilapsn

yep

- 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

you've to treat y as a constant..

- 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

|dw:1442069827280:dw|

- x3_drummerchick

wait i messed that up: it should be

- x3_drummerchick

|dw:1442069945844:dw|

- x3_drummerchick

|dw:1442070088984:dw|

- x3_drummerchick

|dw:1442070113092:dw|

- x3_drummerchick

|dw:1442070133261:dw|

- x3_drummerchick

so now what do i do, and is that correct? ^

- x3_drummerchick

@Jhannybean help :(

- x3_drummerchick

@amilapsn

- amilapsn

c=-y

- x3_drummerchick

i dont think this is right.. can you show some work? that doesnt make sense

- phi

use the quadratic formula, with a=1, b=2, c= -y

- phi

you should get
\[ x= \frac{-2\pm \sqrt{4+4y}}{2} \]
which simplifies to
\[ x= -1 \pm \sqrt{1+y} \]

- 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

so that is the inverse of the function?

- x3_drummerchick

oh okay t hat makes sense

- x3_drummerchick

and so now I just plug in the 15 in for x?

- x3_drummerchick

no wait, dont i have to solve for y first or something?

- 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

ohhhhh okay

- phi

now you have
\[ g^{-1}(x) = -1 + \sqrt{1+x} \]

- x3_drummerchick

thank you so much, that makes alot of sense

- x3_drummerchick

my professor never showed us how to do these kinds -_- totally makes sense though, thanks again!

- phi

the first trick was realizing you use the quadratic formula and treat -y as the "c"

- 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

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

and we cant use -1 because that would make a nonreal number- or imaginary number

- x3_drummerchick

@phi quick question, how did you simplify under the radical??

- x3_drummerchick

what did you do in these 2 steps?|dw:1442072831198:dw|

- x3_drummerchick

|dw:1442072887637:dw|

- phi

4+4y= 4(1+y)
\[ \sqrt{4(1+y) }= \sqrt{4} \sqrt{1+y} \]

- x3_drummerchick

oh wow, okay, thanks!

- 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

so those 2 are completely different, is it because its not held by multiplication?

- phi

you can't split the sum inside a square root.
you are allowed to factor into terms that multiply

- x3_drummerchick

oh okay

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