- anonymous

f(x)=-16x^2 What is the inverse function of.. I need the answer plzzzzzz, assume that all variables are positive. Again, does anyone know the answer? Someone out there is smart, i know there is.......

- schrodinger

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- amistre64

x = -(16y^2)
x/16 = -(y^2)
sqrt(x)/4 = -y
- sqrt(x)/4 = y

- amistre64

maybe

- anonymous

but that would come out to having an i in the final answer correct?

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## More answers

- amistre64

nah, i made sure that I took out the -1 first to avoid that ...maybe lol let me look at it after drawing it

- anonymous

amistre?

- amistre64

yes?

- anonymous

maybe i am tired but i think you missed this one.

- amistre64

oh i miss alot of them ; not just this one lol

- anonymous

ya, this thing is like impossible, it's a question on my test, what should i write down as the answer?

- amistre64

its x = -y^2 but i dont think we can make a function out of it can we?

- anonymous

f says square, then multiply by 16. f inverse says divide by -16, take the square root. should be \[f^{-1}(x)=\sqrt{-\frac{x}{16}}\]

- amistre64

we can stick an absolute value in there cant we :)

- anonymous

one to one because they restricted
\[x>0\]

- amistre64

\[y = \frac{\sqrt{|-x|}}{4}\]

- anonymous

so function is defined only on positive x, and therefore is one to one. range is obviously negative numbers.

- anonymous

i got f^-1=1/4(x) but i know that's wrong......

- anonymous

no range is negative so domain of inverse is negative numbers

- amistre64

\[\frac{\sqrt{|x|}}{4} ; D:(-\infty,0]\]

- amistre64

maybe? lol

- anonymous

i dont have to graph it

- anonymous

absolute value not necessary just put \[\sqrt{\frac{-x}{4}}\]

- amistre64

http://www.wolframalpha.com/input/?i=sqrt%28|x|%29%2F4

- amistre64

http://www.wolframalpha.com/input/?i=sqrt%28%7Cx%7C%29%2F4

- anonymous

ah but that is not the inverse of this function.

- amistre64

gotta run...ciao :)

- anonymous

this function was restricted to positive x.

- anonymous

ok later.

- amistre64

it is if we rectric the domain lol; like trig inverse functions

- anonymous

more puzzles please.

- anonymous

123456789 are you confused yet?

- anonymous

very......i'm a high school freshman in algebra 2 honors.... this is like crazy college crap lol

- anonymous

let me try to xplain. do you know what the graph of \[y=-x^2\] looks like?

- anonymous

yeah, it's a prarabola pointing down

- anonymous

but we arent doing parabloas

- anonymous

well the point is that since it is a parabola it is not a one to one function so it doesn't have an inverse. but they made it a one to one function by restricting the domain to positive numbers. so your actual graph is not a parabola, it is just the right hand side of the parabola. hterefore it is one to one

- anonymous

so the way you do this is write \[y=-16x^2\]
replace y by x and x by y to get \[x=-16y^2\]

- anonymous

ok, now what?

- anonymous

then solve for y:
\[x=-16y^x\]
\[\frac{-x}{16}=y^2\]
\[y=\frac{\sqrt{-x}}{16}\]

- anonymous

again the typesetting is bad. the square root should be over the whole thing.

- anonymous

so it should be \[f^{-1}(x)=\frac{\sqrt{-x}}{4}\]
since the square root of 16 is 4

- anonymous

and you don't have to worry about \[\sqrt{-x}\] being a problem because the range of your original function was negative numbers, so your domain of the inverse is negative numbers.

- anonymous

take out the i or leave it in?

- anonymous

x is negative so -x is positive so you can take its square root.

- anonymous

thanks a bunch

- anonymous

leave it in. the domain of your original function was positive numbers so the range of your inverse must be positive numbers. do not put the minus sign outside. that would make the answer negative.

- anonymous

welcome.

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