1. inverse function help

I dont know if this is right...

i need inverse function of g(x)= -2(x+3)^2 +8 (-5less than or equal to x less than or equal to -3)

so i got... x= -2(y+3)^2 +8
x-8=-2(y+3)^2
-1/2(x-8)= (y+3)^2
square root both sides gives g-1(x)= √ -1/2(x-8)-3
but this seems horribly wrong, - under a root and how do i find the domain from that??

2. First you have to prove that g is bijective, if you do that the problem is like solved.

"so i got... x= -2(y+3)^2 +8
x-8=-2(y+3)^2
-1/2(x-8)= (y+3)^2
square root both sides gives g-1(x)= √ -1/2(x-8)-3"

I don't understand - you should just calculate, then find $\Delta$ (or do that when you prove that g is surjective) and the root that is in the interval you want it to be.

3. Originally Posted by dragon555
I dont know if this is right...

i need inverse function of g(x)= -2(x+3)^2 +8 (-5less than or equal to x less than or equal to -3)

so i got... x= -2(y+3)^2 +8
x-8=-2(y+3)^2
-1/2(x-8)= (y+3)^2
square root both sides gives g-1(x)= √ -1/2(x-8)-3
but this seems horribly wrong, - under a root and how do i find the domain from that??
$g(x) = -2(x+3)^2 +8$ , with domain $-5 \le x \le -3$

range of $g(x)$ is $0 \le y \le 8$

note that the range of $g(x)$ is the domain of $g^{-1}(x)$ ...

$x = -2(y+3)^2 +8$

$x-8 = -2(y+3)^2$

$\dfrac{8-x}{2} = (y+3)^2$

$\pm \sqrt{\dfrac{8-x}{2}} = y+3$

$y = \pm \sqrt{\dfrac{8-x}{2}} - 3$

for the domain $0 \le x \le 8$ , you want the range to be $-5 \le y \le -3$

$g^{-1}(x) = -\sqrt{\dfrac{8-x}{2}} - 3$

4. Originally Posted by dragon555
I dont know if this is right...

i need inverse function of g(x)= -2(x+3)^2 +8 (-5less than or equal to x less than or equal to -3)

so i got... x= -2(y+3)^2 +8
x-8=-2(y+3)^2
-1/2(x-8)= (y+3)^2
square root both sides gives g-1(x)= √ -1/2(x-8)-3
but this seems horribly wrong, - under a root and how do i find the domain from that??
It seesm that this question might be part of an assignment that counts towards your final grade. See rule #6: http://www.mathhelpforum.com/math-he...hp?do=vsarules. Thread closed.