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a)Show that the composition of two one-to-one functions f and g, is one-to-one.
b)Express (f o g)^-1 in terms of f^-1 and g^-1.
For a) I put down:
Assume f and g are one-to-one
g(x1) = x1
g(x2) = x2
f(x1) = x1
f(x2) = x2
g(x1)=g(x2) iff x1=x2
f(x1)=f(x1) iff x1=x2
f(g(x1))=x1
f(g(x2))=x2
f(g(x1))=f(g(x2)) iff x1=x2
Therefore (f o g)(x) are one-to-one
For part b I do not understand how to show it. I know it is true.
it is not the case that g being 1-1 means g(x1) = x1.
for example, g(x) = x+1 is 1-1, but g(x) is never x.
furthermore, you are not being asked to show that (b) is true. you are being asked to give a formula for (f o g)^-1.
what is the definition of 1-1 in your text?
The definition for one-to-one is when the function is able to pass the horizontal line test at all points. However, there is no formula for (f o g)^-1 in forms of f^-1 and g^-1 in my textbook.Quote:
Originally Posted by Deveno
is that what it says in your text, or is that what you have been taught? i ask, because if that is what it says in your text, it is rather a shame. if it is only how 1-1 has been explained to you, then perhaps that can be forgiven.
my reasoning goes like this: "the horizontal line test" is primarily a visual criterion, and using it pre-supposes you have an accurate graph of your function. for complicated functions, it is not intuitively clear how to graph fog. you should have a proper definition of 1-1, which is not limited by one's ability, or inability, to sketch a graph.
a function f is 1-1, IF: whenever f(x1) = f(x2), x1 must equal x2. this captures our intutive feeling that f only maps a single x to every value f(x) (one input, to one output).
using this criterion, we can PROVE that g(x) = x + 1 is 1-1.
suppose that g(x1) = g(x2). that is:
x1 + 1 = x2 + 1. subtracting 1 from both sides:
x1 = x2, so g is 1-1.
this really is the same thing as "the horizontal line test" (f is 1-1 if it doesn't fail the horizontal line test): suppose f has two points on a horizontal line. this means that f(x1) = f(x2), but x1 ≠ x2, so by our formal definition above, f is not 1-1. on the other hand, if f passes the horizontal line test, then the only point on the horizontal line y = f(x1) is (x1,f(x1)), so if f(x2) is on this line, it must be the case that x2 = x1.
what we have done here, is exchange a geometric test, for an analytic test, one which we can check using just algebra, and which is more easily checked just from the formula for f.
