Inverses Flashcards
(14 cards)
Definition of a one to one function:
Consider a f:A->B where f is a function. f is a one to one function iff
For all elements a,b in A, if a is not equal to b then f(a) is not equal to f(b).
Definition of a function’s inverse:
Consider a function f. The inverse of f is the set of all points (b,a) for which the point (a,b) is in f. The inverse of f is denoted f-inverse
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f-inverse is a function iff f is a one to one function.
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The inverse of f-inverse is f
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Consider a function f with domain A and range B.
If f is a one to one function then f-inverse is a one to one function.
The proof uses the fact that the inverse of f-inverse is f
Let f be a one to one function with domain A and range B defined by b = f(a) for all a in A.
Then f- inverse is a (one to one) function with domain B and range A where f-inv(b) = a iff b= f(a)
Then:
For all b in B, f( f-inv(b)) = b
and
For all a in A, f-inv(f(a)) = a.
If f is an increasing function then f is a one to one function.
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If f is an increasing function then f-inverse is increasing
Note: if f is an increasing function then f is a one to one function meaning that f-inverse is a (one to one) function
If f is a continuous, one to one function on an interval, then f is either increasing or decreasing on then interval.
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Let f be a continuous, increasing function with domain [a,b] and range B. Assume that a < b. What can be said about B?
B = [ f(a) , f(b) ] .
Let f be a continuous, increasing function with domain (a,b) and range B. What can be said about B?
B is of the form (-inf, inf) , (- inf, d) , (c,d) or (c, inf) for some c, d in R.
If f is a continuous, one to one function on an open interval then f-inverse is also continuous on the open interval.
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If f is a continuous, one to one function defined on an open interval and f’(f-inv(a)) = 0 then f-inv is not differentiable at a.
If f-inverse was differentiable at a, then use the chain rule to obtain 0 = 1, a contradiction.
Let f be a continuous, one to one function on an interval and suppose f is differentiable at f-inv(b) with f’(f-inv(b)) not equal to 0.
Then:
f- inv is differentiable at b and (f-inv(b))’ = 1/(f’(f-inv(b)))
