FInd the inverse of #f(x)=1+x/3-x# ?

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Having some trouble here, i'd appreciate it if I got some help.

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Answer 1 · 2018-01-11T00:27:18

#f^-1(x)=-3/2(y-1)#

To find the #f^-1(x)# of a function #f(x)=y=....x+....#, we make #x# in terms of #y#.

In here,

#f(x)=1+x/3-x#

#f(x)=1+1/3x-x#

#f(x)=1-2/3x#

Now, just put #y=f(x)#

#y=1-2/3x#

#-2/3x=y-1#

#x=-3/2(y-1)#

Therefore,

#f^-1(x)=-3/2(y-1)#

Answer 2 · 2018-01-11T00:33:41

Given #f(x) =(1+x)/(3-x); x!=3#

Begin by substituting #x = f^-1(x)# into #f(x)#:

#f(f^-1(x)) =(1+f^-1(x))/(3-f^-1(x))#

Use the property #f(f^-1(x)) = x# to make the left side become x:

#x =(1+f^-1(x))/(3-f^-1(x))#

Solve for #f^-1(x)#:

#x(3-f^-1(x))=1+f^-1(x))#

#3x-xf^-1(x)=1+f^-1(x))#

#3x-1-xf^-1(x)=f^-1(x))#

#3x-1=xf^-1(x)+f^-1(x))#

#3x-1=(x+1)f^-1(x))#

#f^-1(x) = (3x-1)/(x+1); x !=-1#

To verify that the above is truly the inverse, you should show that #f(f^-1(x)) = x# and #f^-1(f(x)) = x#