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# composite functions

+1
251
2
+557

f(x)=3-4x

g(x)=3+4x

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

Feb 24, 2018

#1
+18055
+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

Feb 24, 2018

#1
+18055
+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
#2
+99586
+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

Feb 24, 2018