Suppose . Does have an inverse? If so, find . If not, enter "undef".

a)    f(x)   = (9/5)x - 4

Write f(x)   as "y"........the idea is to get x by itself and then "swap" x and y .......so we have...

y = (9/5)x  - 4      add 4 to both sides

y + 4  = (9/5) x    multiply both sides by 5/9

[5 (y + 4) ] / 9    =  x         "swap" x and y

[5 (x + 4)] / 9   = y     for y, write  f^-1(x)  

[5 (x + 4)] / 9  = f^-1(x)

And this is the inverse

So   f^-1(20)  = [5 (20 + 4)] / 9   = [ 120]/ 9

b)  g(x) =   4x^2 + 8x + 13   .....this function is not "one-to-one"..........no inverse exits

c)  h(x)  =   1 / √x     this function is one to one.....but the domain is restricted to x > 0

y = 1/ √x   square both sides

y^2  = 1 / x    "swap"  x and y

x^2  = 1 / y        rearrange as

y =   1/ x^2     for y, write  h^-1(x)

h^-1(x)  = 1/x^2   

So   h^-1(4)   =  1/(4)^2   = 1/16

Note on (b)  above.......I assumed that we were dealing with functions only in this problem......thus my answer that "no inverse exits"  should be amended to "no inverse function exists"  ...however.....

g(x)  = 4x^2 + 8x + 13   has an inverse, but, since the original function is not one-to-one, the "inverse" will not be a function .....the inverse can be derived as follows :

g = 4x^2 + 8x + 13          complete the square  on x

y - 13 + 4  = 4(x^2 + 2x + 1)

[y - 9] / 4  =  (x + 1) ^2         take plus/minus square root of both sides

±sqrt [ y - 9] / 2 = x + 1

±sqrt [ y - 9] / 2  - 1   =  x        "swap" x and y

±sqrt [ x - 9] / 2  - 1   =  y  =  g^-1 (x)

So.....g^-1 (25)   =   ±sqrt [25 - 9]/ 2 - 1   =  ±4/2 - 1   =  ±2 - 1   =   1    and  -3

So  g^-1(25)   returns two values......   1 and -3.......therefore.....it is not a function