anonymous
  • anonymous
Inverse of a Function Help!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Which function below is the inverse of f(x) = x^2 − 9?
anonymous
  • anonymous
The answers have square roots, so I do not know how to solve this, please someone explain!
ankit042
  • ankit042
y = x^2 − 9. Now you have to express x in terms of y

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anonymous
  • anonymous
do you mean switch the x and y variables?
anonymous
  • anonymous
sorry, I'm new to this, haha
anonymous
  • anonymous
\[y = x² - 9\] \[y + 9 = x²\] \[x = \sqrt{y + 9}\]
ankit042
  • ankit042
NO lets take a different example y= 2x+3 then inverse will be (y-3)/2
anonymous
  • anonymous
(1) write down y=f(x) (2) switch x and y to get x=f(y) (3) solve for y to get tex2html_wrap_inline48
anonymous
  • anonymous
this is the guide
anonymous
  • anonymous
where you see an \(x\) put a \(y\) and where you see a \(y\) put an \(x\) it is just crutch, but it helps
anonymous
  • anonymous
suppose you had to solve \[x^2-9=40\] for \(x\) what would you do?
anonymous
  • anonymous
|dw:1377609513071:dw|
anonymous
  • anonymous
step one would be to add \(9\) to both sides to get \(x^2=49\) and step 2 would be to take the two square roots, \(x=\pm\sqrt{49}=\pm7\)
anonymous
  • anonymous
oh my goodness, haha. so many people, thanks everyone! Satellite: So i use the square root whenever i have a variable with an exponent?
anonymous
  • anonymous
that is what the inverse function does add 9, that the two square roots \[\sqrt{x+9}\]
anonymous
  • anonymous
oops i meant \(\pm\sqrt{x+9}\)
anonymous
  • anonymous
yes
anonymous
  • anonymous
@official_snowflake if you want to solve \(x^2=whatever\) then the solution is \(\pm\sqrt{\text{whatever}}\)
anonymous
  • anonymous
but no, you don't always use the square root you might need the cubed root, or the fourth root for example depending on the exponent
anonymous
  • anonymous
oh oh oh okay. I get that!
anonymous
  • anonymous
btw i noticed you wrote in your question "which FUNCTION" below is the inverse of \(f(x)=x^2-9\) but did not include any answers
anonymous
  • anonymous
since the inverse requires \(\pm\) you see it is not a function
anonymous
  • anonymous
would you like the answer options?
anonymous
  • anonymous
for example, since both \(f(7)=40\) and \(f(-7)=40\) if w know only that \(f(x)=40\) we are not sure of what \(x\) is, it might be 7 or it might be -7 that means the inverse cannot be a function
anonymous
  • anonymous
the inverse function is
anonymous
  • anonymous
sure i would love to
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
|dw:1377610057169:dw|
anonymous
  • anonymous
the first answer is supposed to be correct, and the person who wrote the problem doesn't understand what a function is too bad no wonder students get confused
anonymous
  • anonymous
will add this to my "bad math" collection it is not your fault you are being taught by the less than competent
anonymous
  • anonymous
oh! haha, that's crazy. So non of those are functions?
anonymous
  • anonymous
there is no such "function" as \(\pm\sqrt{x+9}\)
anonymous
  • anonymous
the second and 4th are functions the first and third are not
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Inverse_function
anonymous
  • anonymous
the reason is simple, so don't let these morons confuse you a function has to be well defined meaning that if plug in a number and i plug in a number, we must both get exactly the same number out there is no room for ambiguity
anonymous
  • anonymous
oh weird. okay, that goes for every function?
anonymous
  • anonymous
the "plus minus" at the beginning of 1 and 3 means that you do not have a functions could be plus... could be minus functions do not tolerate that ambiguity
anonymous
  • anonymous
inverse function nga e sipa in kita wla ka pala sa akin ...mali lahat ng sagot mo
anonymous
  • anonymous
\[f(x)=x^2\] is a function for sure plug in a number, get out only one number but suppose i try to solve \(x^2=9\) for \(x\) i only know that either \(x=3\) or \(x=-3\) so i have two choices that means \(\pm\sqrt{x}\) is NOT a function
anonymous
  • anonymous
okay, i think I get that. Do you mind if I ask you one more question? I don't mean to bother you..
anonymous
  • anonymous
this is the answer f^-1= square root of x^+9 .!!!! f^-1 means f inverse function of
anonymous
  • anonymous
sure go ahead and ask
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Inverse_function#Formula_for_the_inverse
anonymous
  • anonymous
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anonymous
  • anonymous
i thought i understood it, then I saw the answers to choose from and I have no idea how they went about solving this!
anonymous
  • anonymous
it just looks like they flipped random numbers around!
anonymous
  • anonymous
ok before we do this lets solve this problem \[\frac{4x-3}{2}=13\] what would you do?
anonymous
  • anonymous
would i start with adding 3?
anonymous
  • anonymous
i wouldn't the left sides is a fraction
anonymous
  • anonymous
if it was \(4x-3=13\) i would add 3, but it is \(\frac{4x-3}{2}=13\)
anonymous
  • anonymous
oh so would i multiply two?
anonymous
  • anonymous
yes
anonymous
  • anonymous
okay, sorry. that was obvious haha
anonymous
  • anonymous
@satellite73 , how do you do inline equations?
anonymous
  • anonymous
lets write the steps 1) multiply by 2, that gives \(4x-3=26\)
anonymous
  • anonymous
\(
anonymous
  • anonymous
THEN you would add 3
anonymous
  • anonymous
:)
anonymous
  • anonymous
right
anonymous
  • anonymous
1) multiply by 2 2) add 3 then what?
anonymous
  • anonymous
then you divide by 4
anonymous
  • anonymous
ok we 1) multiply by 2 2) add 3 3) divide by 4 now lets write that as a function
anonymous
  • anonymous
@Anickyan if you ever want to see code, right click and select "show math as" then "latex"
anonymous
  • anonymous
to put equations in line, use \( instead of \[ you can do that in regular latex as well, works like $
anonymous
  • anonymous
ok now we write our inverse function 1) multiply by 2 2) add 3 3) divide by 4
anonymous
  • anonymous
take \(x\) then \(2x\) then \(2x+3\) then \(\frac{2x+3}{4}\)
anonymous
  • anonymous
so \[f^{-1}(x)=\frac{2x+3}{4}\]
anonymous
  • anonymous
Ohh, I get it. You sort of have to solve it first and put it back together. (: thank you so much.
anonymous
  • anonymous
or we can do this a different way put \[x=\frac{4y-3}{2}\] and solve for \(y\)
anonymous
  • anonymous
but you should see it is exactly the same \[x=\frac{4y-3}{2}\] \[2x=4y-3\] \[2x+3=4y\] \[\frac{2x+3}{4}=y\]
anonymous
  • anonymous
you should learn the second method because sometimes the first method won't work

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