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Given f(x) = 2/3(x-5), determine the value of the following:

a) f^-1(-2)                   b) f^-1(a+1) - f^-1(a)

 

c) [ f^-1(a)-f^-1(b) ] / a-b

 

the process is most important to me.. please don't use anything more complicated than inverse funtions

also all the f^-1 are inverse functions

THANKS!!

 Jan 9, 2019
 #1
avatar+98154 
+2

Let's write  y = (2/3) ( x - 5)       multiply both sides by 3/2

 

(3/2)y =  x - 5       add 5 to both sides

 

(3/2)y + 5 =  x        "swap" x and y

 

(3/2)x + 5   =  y  = f-1(x)       this is the inverse

 

 

 

So

 

a) f-1 (-2)   =  (3/2)(-2) + 5  =  - 3 + 5  =  2

 

b)  f-1 ( a + 1) - f-1 ( a)  =   [ (3/2)(a + 1)   + 5 ] - [ (3/2)a + 5 ]  = 3/2

 

c )  [  f-1(a)  - f-1(b) ] / [a - b]  =  (  [ (3/2)(a + 5]  -  (3/2)(b) + 5] / [ a - b]  =

 

[ (3/2) (a - b) ] / [ a - b ]  =    3/2

 

 

 

cool cool cool

 Jan 9, 2019
 #2
avatar+63 
+1

Hey CPhill, thanks for the help!

I still have one question based on your work : )

Here is the work:

[  f-1(a)  - f-1(b) ] / [a - b]  =  (  [ (3/2)(a + 5]  -  (3/2)(b) + 5] / [ a - b]  =

 

[ (3/2) (a - b) ] / [ a - b ]  =    3/2

 

I don't understand the bolded out part, as in how you went from [ (3/2)(a + 5]  -  (3/2)(b) + 5] / [ a - b] to the next part..

Could you please further explain your steps, I am a little confused about that

Drazil  Jan 9, 2019
 #3
avatar+98154 
+2

OK.....let's take it from here

 

( [ (3/2) (a) +  5 ]  -  [ (3/2) (b) + 5 ] )  / [ a - b ]  =

 

[ (3/2)a + 5 - (3/2)b - 5 ]  / [ a - b ]  =

 

[ (3/2) a - (3/2)(b)] / [ a - b]  =

 

 (3/2) (a - b) / ( a - b)  =

 

3/2

 

I think I forgot to close the parentheses  on the (a) term above....sorry....!!! 

 

cool cool cool

CPhill  Jan 9, 2019
edited by CPhill  Jan 9, 2019
edited by CPhill  Jan 9, 2019
 #4
avatar+63 
+1

Yeah, I was a little confused. All good, thanks!

Drazil  Jan 9, 2019
 #5
avatar+98154 
0

No prob....!!!

 

 

cool cool cool

CPhill  Jan 9, 2019

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