+115 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!!
+128475 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
+115 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
+128475 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....!!!
