Calc Question: one-to-one functions and inverses?

Question

a) 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

http://www.analyzemath.com/OneToOneFunct...

b) f(x)=1/(x-3)+3

y=f(x)

y=1/(x-3)+3

• Switch x and y ==> x=1/(y-3)+3

•• Solve for y=g(x) ==>

x-3=1/(y-3)

y-3=1/(x-3)

y=1/(x-3)+3

g(x)=f^-1(x)=f(x)

c) If a function h is its own inverse, then the bisector of the first and the third quadrant y=x is an axis of symmetry of the graph of y=h(x).