+0  
 
+1
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avatar+549 

f(x)=3-4x

g(x)=3+4x

 

show that  f^-1(x)+g^-1(x)=0, for all the values of x.

 

winkwinkwink

lynx7  Feb 24, 2018

Best Answer 

 #1
avatar+13577 
+2

The first one    f(x) = y = 3-4x    solve for 'x'

y-3 = -4x

3-y = 4x

(3-y)/4 = x      now switch the x's and y's

f^-1 (x) =  (3-x)/4

 

The second one  y=3+4x

                            (y-3)/4 = x

f^-1 (x) =  (x-3)/4

 

Now add the two f^-1  (x) in red       (3-x)/4 + (x-3)/4    =  (x-x +3-3)/4 = 0

ElectricPavlov  Feb 24, 2018
 #1
avatar+13577 
+2
Best Answer

The first one    f(x) = y = 3-4x    solve for 'x'

y-3 = -4x

3-y = 4x

(3-y)/4 = x      now switch the x's and y's

f^-1 (x) =  (3-x)/4

 

The second one  y=3+4x

                            (y-3)/4 = x

f^-1 (x) =  (x-3)/4

 

Now add the two f^-1  (x) in red       (3-x)/4 + (x-3)/4    =  (x-x +3-3)/4 = 0

ElectricPavlov  Feb 24, 2018
 #2
avatar+90968 
+2

Find the inverses

 

y =  3 - 4x                                         y   =   3 + 4x                      

y - 3  = - 4x                                       y - 3  =  4x

3 - y  = 4x                                         [ y - 3 ] / 4  = x

[3 - y ] / 4  = x                                   [ x -  3 ] / 4  =  the inverse

[3 - x] / 4  = the inverse

 

So..the sum of these inverses is :

 

[ 3 - x] / 4   +  [ x - 3] / 4  =

 

[ 3 - 3 - x + x ] / 4   =    0 / 4   =   0

 

 

 

cool cool cool

CPhill  Feb 24, 2018

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