Question

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.......

Answer 1

x = -(16y^2)

x/16 = -(y^2)

sqrt(x)/4 = -y

- sqrt(x)/4 = y

Answer 2

maybe

Answer 3

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

Answer 4

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

Answer 5

amistre?

Answer 6

yes?

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

no range is negative so domain of inverse is negative numbers

Answer 18

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

Answer 19

maybe? lol

Answer 20

i dont have to graph it

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

ah but that is not the inverse of this function.

Answer 25

gotta run...ciao :)

Answer 26

this function was restricted to positive x.

Answer 27

ok later.

Answer 28

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

Answer 29

more puzzles please.

Answer 30

123456789 are you confused yet?

Answer 31

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

Answer 32

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

Answer 33

yeah, it's a prarabola pointing down

Answer 34

but we arent doing parabloas

Answer 35

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

Answer 36

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

Answer 37

ok, now what?

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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.

Answer 42

take out the i or leave it in?

Answer 43

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

Answer 44

thanks a bunch

Answer 45

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.

Answer 46

welcome.