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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!!

Drazil Jan 9, 2019

#1**+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

CPhill Jan 9, 2019

#2**+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