anonymous
  • 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.......
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
x = -(16y^2) x/16 = -(y^2) sqrt(x)/4 = -y - sqrt(x)/4 = y
amistre64
  • amistre64
maybe
anonymous
  • anonymous
but that would come out to having an i in the final answer correct?

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More answers

amistre64
  • 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
  • anonymous
amistre?
amistre64
  • amistre64
yes?
anonymous
  • anonymous
maybe i am tired but i think you missed this one.
amistre64
  • amistre64
oh i miss alot of them ; not just this one lol
anonymous
  • anonymous
ya, this thing is like impossible, it's a question on my test, what should i write down as the answer?
amistre64
  • amistre64
its x = -y^2 but i dont think we can make a function out of it can we?
anonymous
  • 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
  • amistre64
we can stick an absolute value in there cant we :)
anonymous
  • anonymous
one to one because they restricted \[x>0\]
amistre64
  • amistre64
\[y = \frac{\sqrt{|-x|}}{4}\]
anonymous
  • anonymous
so function is defined only on positive x, and therefore is one to one. range is obviously negative numbers.
anonymous
  • anonymous
i got f^-1=1/4(x) but i know that's wrong......
anonymous
  • anonymous
no range is negative so domain of inverse is negative numbers
amistre64
  • amistre64
\[\frac{\sqrt{|x|}}{4} ; D:(-\infty,0]\]
amistre64
  • amistre64
maybe? lol
anonymous
  • anonymous
i dont have to graph it
anonymous
  • anonymous
absolute value not necessary just put \[\sqrt{\frac{-x}{4}}\]
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=sqrt%28|x|%29%2F4
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=sqrt%28%7Cx%7C%29%2F4
anonymous
  • anonymous
ah but that is not the inverse of this function.
amistre64
  • amistre64
gotta run...ciao :)
anonymous
  • anonymous
this function was restricted to positive x.
anonymous
  • anonymous
ok later.
amistre64
  • amistre64
it is if we rectric the domain lol; like trig inverse functions
anonymous
  • anonymous
more puzzles please.
anonymous
  • anonymous
123456789 are you confused yet?
anonymous
  • anonymous
very......i'm a high school freshman in algebra 2 honors.... this is like crazy college crap lol
anonymous
  • anonymous
let me try to xplain. do you know what the graph of \[y=-x^2\] looks like?
anonymous
  • anonymous
yeah, it's a prarabola pointing down
anonymous
  • anonymous
but we arent doing parabloas
anonymous
  • 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
  • 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
  • anonymous
ok, now what?
anonymous
  • anonymous
then solve for y: \[x=-16y^x\] \[\frac{-x}{16}=y^2\] \[y=\frac{\sqrt{-x}}{16}\]
anonymous
  • anonymous
again the typesetting is bad. the square root should be over the whole thing.
anonymous
  • anonymous
so it should be \[f^{-1}(x)=\frac{\sqrt{-x}}{4}\] since the square root of 16 is 4
anonymous
  • 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
  • anonymous
take out the i or leave it in?
anonymous
  • anonymous
x is negative so -x is positive so you can take its square root.
anonymous
  • anonymous
thanks a bunch
anonymous
  • 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
  • anonymous
welcome.

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