Inverse of a Function Help!

- anonymous

Inverse of a Function Help!

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- anonymous

Which function below is the inverse of f(x) = x^2 − 9?

- anonymous

The answers have square roots, so I do not know how to solve this, please someone explain!

- ankit042

y = x^2 − 9. Now you have to express x in terms of y

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## More answers

- anonymous

do you mean switch the x and y variables?

- anonymous

sorry, I'm new to this, haha

- anonymous

\[y = x² - 9\]
\[y + 9 = x²\]
\[x = \sqrt{y + 9}\]

- ankit042

NO lets take a different example y= 2x+3 then inverse will be (y-3)/2

- 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

this is the guide

- 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

suppose you had to solve
\[x^2-9=40\] for \(x\)
what would you do?

- anonymous

|dw:1377609513071:dw|

- 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

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

that is what the inverse function does
add 9, that the two square roots
\[\sqrt{x+9}\]

- anonymous

oops i meant \(\pm\sqrt{x+9}\)

- anonymous

yes

- anonymous

@official_snowflake if you want to solve \(x^2=whatever\) then the solution is \(\pm\sqrt{\text{whatever}}\)

- 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

oh oh oh okay. I get that!

- 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

since the inverse requires \(\pm\) you see it is not a function

- anonymous

would you like the answer options?

- 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

the inverse function is

- anonymous

sure i would love to

- anonymous

##### 1 Attachment

- anonymous

|dw:1377610057169:dw|

- 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

will add this to my "bad math" collection
it is not your fault you are being taught by the less than competent

- anonymous

oh! haha, that's crazy. So non of those are functions?

- anonymous

there is no such "function" as \(\pm\sqrt{x+9}\)

- anonymous

the second and 4th are functions
the first and third are not

- anonymous

http://en.wikipedia.org/wiki/Inverse_function

- 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

oh weird. okay, that goes for every function?

- 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

inverse function nga e sipa in kita wla ka pala sa akin ...mali lahat ng sagot mo

- 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

okay, i think I get that. Do you mind if I ask you one more question? I don't mean to bother you..

- anonymous

this is the answer f^-1= square root of x^+9 .!!!! f^-1 means f inverse function of

- anonymous

sure go ahead and ask

- anonymous

http://en.wikipedia.org/wiki/Inverse_function#Formula_for_the_inverse

- anonymous

##### 1 Attachment

- 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

it just looks like they flipped random numbers around!

- anonymous

ok before we do this lets solve this problem
\[\frac{4x-3}{2}=13\] what would you do?

- anonymous

would i start with adding 3?

- anonymous

i wouldn't
the left sides is a fraction

- anonymous

if it was \(4x-3=13\) i would add 3, but it is \(\frac{4x-3}{2}=13\)

- anonymous

oh so would i multiply two?

- anonymous

yes

- anonymous

okay, sorry. that was obvious haha

- anonymous

@satellite73 , how do you do inline equations?

- anonymous

lets write the steps 1) multiply by 2, that gives \(4x-3=26\)

- anonymous

\(

- anonymous

THEN you would add 3

- anonymous

:)

- anonymous

right

- anonymous

1) multiply by 2
2) add 3
then what?

- anonymous

then you divide by 4

- anonymous

ok we
1) multiply by 2
2) add 3
3) divide by 4
now lets write that as a function

- anonymous

@Anickyan if you ever want to see code, right click and select "show math as" then "latex"

- anonymous

to put equations in line, use \( instead of \[
you can do that in regular latex as well, works like $

- anonymous

ok now we write our inverse function
1) multiply by 2
2) add 3
3) divide by 4

- anonymous

take \(x\) then \(2x\) then \(2x+3\) then \(\frac{2x+3}{4}\)

- anonymous

so
\[f^{-1}(x)=\frac{2x+3}{4}\]

- anonymous

Ohh, I get it. You sort of have to solve it first and put it back together. (: thank you so much.

- anonymous

or we can do this a different way
put
\[x=\frac{4y-3}{2}\] and solve for \(y\)

- 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

you should learn the second method because sometimes the first method won't work

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