1. INVERSE FUNCTIONS (reversing the process) Consider the following three functions f: A  B A B abcabc defdef abcabc defdef abcabc defdef Suppose we now reverse the arrows to obtain relationships from B  A. Are these still functions ?

2. B A abcabc defdef abcabc defdef abcabc defdef Not a function! A function.Not a function! The reason that the second example is also a function in reverse is because there is a one to one correspondence between the elements.

3. Note: If f: A  B and g: B  A with x  A and y  B andf(x) = y & g(y) = x ie ABAB x y f g Then we say that g is the inverse function of f and we write this as f -1

4. In simple terms Function-gives y in terms of x. Inverse-gives x in terms of y. However inverse formula is written using x – not y ! NB ABBAABBA abcabc defdef abcabc defdef f f -1 Domain of f = {a, b, c} = Range of f -1 Domain of f -1 = {d, e, f} = Range of f

5. Ex f:R  R is given by f(x) = 3x – 2. Find f -1. f(x) = 3x – 2 ******** y = 3x - 2Want x in terms of y ! 3x = y + 2 x = 1 / 3 (y + 2) Formula must be given using x ! f -1 (x) = 1 / 3 (x + 2) f -1 (y) = 1 / 3 (y + 2)

6. Ex2 f:R  R is defined by the formula f(x) = 1 – x 2x + 1 (i) Find a suitable domain for f. (ii) Find a formula for f -1. (iii) Find a suitable domain for f -1 and hence the range of f. ******* (i)2x +1  0so 2x  -1so x  -1 / 2 Hence domain = {x  R: x  -1 / 2 }

7. (ii) f(x) = 1 – x 2x + 1 y = 1 – x 2x + 1 y = (1 – x) 1 (2x + 1) y(2x + 1 )= 1(1 – x) 2xy + y = 1 – x 2xy + x = 1 – y x(2y + 1 )= (1 – y) x = 1 – y 2y + 1 f -1 (x) = 1 – x 2x + 1 = f(x) coincidence (iii) hence Domain of f -1 = Range of f = {x  R: x  -1 / 2 }