Unformatted text preview: k −1 or f (a + k ) = b + h. (b) = a and f (a) = b, by (0.2) and (0.1), we get
a+k−a
(f −1 ) (b) = lim
h→ 0
h
k
= lim
.
h→0 f (a + k ) − f (a) Notice that, as h approaches zero, k = f −1 (b + h) − a also approaches zero, since
lim (f −1 (b + h) − a) = f −1 (lim (b + h)) − a = f −1 (b) − a = 0, h→ 0 h→ 0 where we have used the continuity of f −1 in the ﬁrst equality above. Note that f −1 is
continuous, since f is continuous. Therefore, we can rewrite (0.3) as
k
(f −1 ) (b) = lim
k → 0 f ( a + k ) − f ( a)
1
f ( a + k ) − f ( a)
=
since lim
= f ( a) ,
k →0
f ( a)
k
1
= −1
.
f (f (b))
1...
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This note was uploaded on 02/10/2014 for the course MATH 235 taught by Professor Lucas during the Fall '08 term at James Madison University.
 Fall '08
 LUCAS
 Calculus, Derivative

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