INverse Fuction help (1 Viewer)

evilevoevil

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Need help with these inverse function questions

Find points of intersection of curves y = f(x) and y = inverse f )x_ if


f(x) = (x-1)^2 , x>1

f(x) = 3x (2-x), x < or equal to 1

Teacher didn't explain it very well so im kinda lost
 

annabackwards

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The points of intersections of inverse functions occurs at the line y = x (because an inverse graph is practically a mirror of the original graph about the line y = x).

So just solve the equations above simultaneously with the line y = x and pick the answers that satisfy the given domain :)
 

evilevoevil

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Thanks.

Also, if it asks for you to find the inverse function of each function and verify that f-1 [f(x)] = f[f-1 (x)] =x, how do you do that? I dont understand how you do the verify bit.
 

oly1991

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thats just provinf there mutually inverse

e.g.

f(x)=x^3 , therefore f^-1(x)=(3root)x
f^-1[f(x)]=f^-1(x^3)
=(3root)(x^3)
=x

f[f^-1(x)]=f((3root)x)
=((3root)x^3
=x

hope that helps
 

evilevoevil

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The points of intersections of inverse functions occurs at the line y = x (because an inverse graph is practically a mirror of the original graph about the line y = x).

So just solve the equations above simultaneously with the line y = x and pick the answers that satisfy the given domain :)
When solving simultaneously, does the point/s have to satisfyboth the function and inverse function domains and ranges? I'm a little confused at this point.

thats just provinf there mutually inverse

e.g.

f(x)=x^3 , therefore f^-1(x)=(3root)x
f^-1[f(x)]=f^-1(x^3)
=(3root)(x^3)
=x

f[f^-1(x)]=f((3root)x)
=((3root)x^3
=x

hope that helps
So you have to show both steps to get the marks?
 

annabackwards

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When solving simultaneously, does the point/s have to satisfy both the function and inverse function domains and ranges? I'm a little confused at this point.
Only for the domain and/or range of f(x) is necessary as it'll satisfy the fomain of the inverse function anyway.
 

Aquawhite

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Hi. I'm y=x :)....

I'm very special. If you don't use me in inverse functions, you won't get good marks.

hehe.
 

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