Calculus Inverse Functions October 3, 2018
Definitions
(a) A function f is said to be one-to-one if there are no two distinct numbers in the domain of f at which f takes the same value, i.e.
if f (x1) = f (x2) then x1 = x2.
(b) Fact If f is a one-to-one function, then there is one and only one function f defined on the range of f that safisfies the equation
f (g(x)) = x for all x in the range of f.
(c) Let f be a one-to-one function. The inverse of f, denoted by f−1, is the unique function defined on the range of f that satisfies the equation
f (f−1(x)) = x for all x in the range of f.
(d) Facts Let f be a one-to-one function. Then (i) f−1(f (x)) = x for all x in the domain of f.
(ii) domain of f−1 = range of f and range of f−1 = domain of f.
(iii) the graph of f−1 is the graph of f reflected in the line y = x.
Theorem (Continuity of the Inverse) Let f be a one-to-one function defined on an interval (a, b). If f is continuous, then its inverse f−1 is also continuous.
Remarks
(a) Let f be a one-to-one continuous function defined on (a, b). Then f either increases through- out (a, b) or it decreases throughout (a, b).
Proof Suppose that f is neither increasing nor decreasing throughout (a, b) and there exist x1 < x2 < x3 in (a, b) such that
f (x1) < f (x2) and f (x1) < f (x3) < f (x2).
Since f is continuous on [x1, x2], by the Intermediate Value Theorem, there exists c ∈ (x1, x2) such that c 6= x3, f (c) = f (x3).
This is a contradiction since f is one-to-one on (a, b).
(b) Let f : (a, b) → (c, d) be a one-to-one function from (a, b) onto (c, d). Suppose that f is increasing on (a, b) then f−1 is increasing on (c, d).
Proof Since f is increasing on (a, b) and f−1: (c, d) → (a, b) is onto, for any y1 < y2 ∈ (c, d) there exist x1, x2 ∈ (a, b) such that
f (x1) = y1 < y2 = f (x2) =⇒ f−1(y1) = x1 < x2 = f−1(y2).
This proves that f−1 is increasing on (c, d).
Calculus Inverse Functions (Continued) October 3, 2018
Proof If f is continuous, then, being one-to-one, f either increases throughout (a, b) or it de- creases throughout (a, b).
Suppose now that f increases throughout (a, b). Let’s take c in the domain of f−1 and show that f−1 is continuous at c.
We first observe that f−1(c) lies in (a, b) and choose > 0 sufficiently small so that f−1(c) − and f−1(c) + also lie in (a, b), i.e.
(f−1(c) − , f−1(c) + ) ⊂ (a, b).
Choose δ > 0 such that
f (f−1(c) − ) < c − δ and c + δ < f (f−1(c) + ), i.e.
(c − δ, c + δ) ⊂ (f (f−1(c) − ), f (f−1(c) + )).
If c − δ < x < c + δ, then
f (f−1(c) − ) < c − δ < x < c + δ < f (f−1(c) + ) =⇒ f (f−1(c) − ) < x < f (f−1(c) + ), and, since f−1 also increases,
f−1(c) − < f−1(x) < f−1(c) + .
The case where f decreases throughout (a, b) can be handled in a similar manner.
if c − δ < x < c + δ, then f−1(c) − < f−1(x) < f−1(c) + .
This condition can be met by choosing δ to satisfy
Theorem (Differentiability of the Inverse) Let f be a one-to-one function differentiable on an open interval I. Let a be a point of I and let f (a) = b. If f0(a) 6= 0, then f−1 is differentiable at b and
(f−1)0(b) = 1 f0(a).
Proof Take > 0,since f is differentiable at a, exists a δ1 > 0 such that
if 0 < |x − a| < δ1, then
1 f (x) − f (a)
x − a
− 1
f0(a)
< ,
and therefore
x − f−1(b) f (x) − b − 1
f0(a)
=
x − a
f (x) − f (a) − 1 f0(a)
< .
Also since f is continuous at a, f−1 is continuous at b. Hence there exists a δ > 0 such that if 0 < |t − b| < δ then 0 < |f−1(t) − f−1(b)| < δ1,
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Calculus Inverse Functions (Continued) October 3, 2018 and therefore
f−1(t) − f−1(b)
t − b − 1
f0(a)
< .
This shows that f−1 is differentiable at b and
(f−1)0(b) = 1 f0(a).
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