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Mathematics 17 Online
OpenStudy (error1603):

Please Help will give medal

OpenStudy (error1603):

How do I find the inverse function of this, and check my answer? I need a detailed explanation please

OpenStudy (error1603):

OpenStudy (brooke2353):

well first what are the steps for finding an inverse operation??

OpenStudy (error1603):

umm

OpenStudy (brooke2353):

1. First, replace with y. This is done to make the rest of the process easier. 2. Replace every x with a y and replace every y with an x. 3. Solve the equation from Step 2 for y. ... 4. Replace y with . ... 5. Verify your work by checking that and are both true.

OpenStudy (error1603):

x = (y^2 + 3)/(3y^2)

OpenStudy (brooke2353):

there you go

OpenStudy (error1603):

So thats it?

OpenStudy (brooke2353):

yeah thats the steps to solving that equation so lets say.. Here's how the process works: Here's my original function: y = 3x - 2 Now I'll try to solve for "x =": y + 2 = 3x (y + 2)/3 = x Once I have "x =", I'll switch x and y; the "y =" is the inverse. y = (x + 2)/3

OpenStudy (error1603):

then x=2 gives y=7/12

OpenStudy (brooke2353):

perfect

OpenStudy (error1603):

so sub x=7/12 into y=sqrt(3/(3x-1)

OpenStudy (brooke2353):

uhhh... what?

OpenStudy (brooke2353):

oh okay my bad i was lost for a second

OpenStudy (error1603):

Wait im lost

OpenStudy (brooke2353):

lol i forgot to plug the square rool into the equation you were right

OpenStudy (error1603):

sub x=(1/3 + 1/x^2) into sqrt(3/(3x-1)?

OpenStudy (brooke2353):

yup

OpenStudy (brooke2353):

and then so forth

OpenStudy (error1603):

So f^-1(x) = √(3/(3x-1))

OpenStudy (error1603):

Okay thanks

OpenStudy (brooke2353):

*clap clap* thats the final answer

OpenStudy (error1603):

Thanks alot

OpenStudy (brooke2353):

mhm

OpenStudy (brooke2353):

its no problem

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