14.6. INDEPENDENCE OF PARAMETRIZATION∗ 271
Proof: Since f is either strictly increasing or strictly decreasing, it follows that f (a,b)is an open interval (c,d). Assume f is decreasing. Now let x∈ (a,b). Why is f−1 is contin-uous at f (x)? Since f is decreasing, if f (x)< f (y), then y≡ f−1 ( f (y))< x≡ f−1 ( f (x))and so f−1 is also decreasing. Let ε > 0 be given. Let ε >η > 0 and (x−η ,x+η)⊆ (a,b).Then f (x) ∈ ( f (x+η) , f (x−η)). Let
δ = min( f (x)− f (x+η) , f (x−η)− f (x)) .
Then if | f (z)− f (x)|< δ , it follows
z≡ f−1 ( f (z)) ∈ (x−η ,x+η)⊆ (x− ε,x+ ε)
which implies ∣∣ f−1 ( f (z))− x∣∣= ∣∣ f−1 ( f (z))− f−1 ( f (x))
∣∣< ε.
This proves the theorem in the case where f is strictly decreasing. The case where f isincreasing is similar. ■
Theorem 14.6.3 Let f : [a,b]→ R be continuous and one to one. Suppose f ′ (x1) existsfor some x1 ∈ [a,b] and f ′ (x1) ̸= 0. Then
(f−1)′( f (x1)) exists and is given by the formula(
f−1)′( f (x1)) =
1f ′(x1)
.
Proof: By Lemma 14.6.1 f is either strictly increasing or strictly decreasing and f−1 iscontinuous on [a,b]. Therefore there exists η > 0 such that if 0 < | f (x1)− f (x)|< η , then
0 < |x1− x|=∣∣ f−1 ( f (x1))− f−1 ( f (x))
∣∣< δ
where δ is small enough that for 0 < |x1− x|< δ ,∣∣∣∣ x− x1
f (x)− f (x1)− 1
f ′ (x1)
∣∣∣∣< ε.
It follows that if 0 < | f (x1)− f (x)|< η ,∣∣∣∣ f−1 ( f (x))− f−1 ( f (x1))
f (x)− f (x1)− 1
f ′ (x1)
∣∣∣∣= ∣∣∣∣ x− x1
f (x)− f (x1)− 1
f ′ (x1)
∣∣∣∣< ε
Therefore, since ε > 0 is arbitrary,
limy→ f (x1)
f−1 (y)− f−1 ( f (x1))
y− f (x1)=
1f ′ (x1)
. ■
The following obvious corollary comes from the above by not bothering with endpoints.
Corollary 14.6.4 Let f : (a,b)→ R be continuous and one to one. Suppose f ′ (x1) existsfor some x1 ∈ (a,b) and f ′ (x1) ̸= 0. Then
(f−1)′( f (x1)) exists and is given by the formula(
f−1)′( f (x1)) =
1f ′(x1)
.
Proof: From the definition of the derivative and continuity of f−1,
limf (x)→ f (x1)
f−1 ( f (x))− f−1 ( f (x1))
f (x)− f (x1)= lim
x→x1
x− x1
f (x)− f (x1)=
1f ′ (x1)
. ■