One-to-one Functions and Inverses?

Question

) let f be a one-to-one function and let g(x)= f^-1 (x). Can you be sure that g is one-to-one? Explain.

b) Show that f(x)= [1/ (x-3)] +3 is its own inverse.

c) Ifa function h is its own inverse, what can you say about the graph of y= h(x)?

Any help on this problem would be appreciated. Please explain throroughly so that I can fully understand and do similar problems on my own. Thanks so much and 10 points to the best answer!

Answer 1

a)  yes, if f(x) is 1-1, then f^-1 (x) is also 1-1

in a 1-1 function, neither the x- nor y-coordinates are repeated

therefore, when you switch x and y for the inverse, neither will still be repeated

a 1-1 passes both the VLT and the HLT; so will its inverse

b)  f(x) = y = 1 / (x - 3) + 3

remember to find the inverse, you switch x and y, then solve for y

inverse: x = 1 / (y - 3) + 3

x - 3 = 1 / (y  - 3)

(x - 3)(y - 3) = 1

y - 3 = 1( / x - 3)

y = 1 / (x - 3) + 3

therefore, since y = f^-1(x) = f(x), the function it it's own inverse

check: f(4) = 1/1 + 3 = 4

g(4) = 1/1 + 3 = 4

so (4,4) maps into (4,4)

f(5) = 1/2 + 3 = 7/2

g(7/2) = 1 /(1/2) + 3 = 5

(5,7/2) maps into (7/2, 5)

this checks for definition of inverse (a,b) ==> (b,a)

c)a function and its inverse are reflected about the line y = x

if a function is its own inverse, that means that it is symmetric about the line y = x