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
