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Chapter 8
Differentiable Functions
A differentiable function is a function that can be approximated locally by a linear function.
8.1. The derivative
Definition 8.1. Suppose that f : (a, b) → R and a < c < b. Then f is differentiable at c with derivative f ′(c) if
lim h→ 0
[
f (c + h) − f (c) h
]
= f ′(c).
The domain of f ′^ is the set of points c ∈ (a, b) for which this limit exists. If the limit exists for every c ∈ (a, b) then we say that f is differentiable on (a, b).
Graphically, this definition says that the derivative of f at c is the slope of the tangent line to y = f (x) at c, which is the limit as h → 0 of the slopes of the lines through (c, f (c)) and (c + h, f (c + h)).
We can also write
f ′(c) = lim x→c
[
f (x) − f (c) x − c
]
since if x = c + h, the conditions 0 < |x − c| < δ and 0 < |h| < δ in the definitions of the limits are equivalent. The ratio
f (x) − f (c) x − c
is undefined (0/0) at x = c, but it doesn’t have to be defined in order for the limit as x → c to exist.
Like continuity, differentiability is a local property. That is, the differentiability of a function f at c and the value of the derivative, if it exists, depend only the values of f in a arbitrarily small neighborhood of c. In particular if f : A → R
140 8. Differentiable Functions
where A ⊂ R, then we can define the differentiability of f at any interior point c ∈ A since there is an open interval (a, b) ⊂ A with c ∈ (a, b).
8.1.1. Examples of derivatives. Let us give a number of examples that illus- trate differentiable and non-differentiable functions.
Example 8.2. The function f : R → R defined by f (x) = x^2 is differentiable on R with derivative f ′(x) = 2x since
lim h→ 0
[
(c + h)^2 − c^2 h
]
= lim h→ 0
h(2c + h) h
= lim h→ 0 (2c + h) = 2c.
Note that in computing the derivative, we first cancel by h, which is valid since h 6 = 0 in the definition of the limit, and then set h = 0 to evaluate the limit. This procedure would be inconsistent if we didn’t use limits.
Example 8.3. The function f : R → R defined by
f (x) =
x^2 if x > 0 , 0 if x ≤ 0.
is differentiable on R with derivative
f ′(x) =
2 x if x > 0 , 0 if x ≤ 0.
For x > 0, the derivative is f ′(x) = 2x as above, and for x < 0, we have f ′(x) = 0. For 0, we consider the limit
lim h→ 0
[
f (h) − f (0) h
]
= lim h→ 0
f (h) h
The right limit is
lim h→ 0 +
f (h) h
= lim h→ 0
h = 0,
and the left limit is
lim h→ 0 −
f (h) h
Since the left and right limits exist and are equal, the limit also exists, and f is differentiable at 0 with f ′(0) = 0.
Next, we consider some examples of non-differentiability at discontinuities, cor- ners, and cusps.
Example 8.4. The function f : R → R defined by
f (x) =
1 /x if x 6 = 0, 0 if x = 0,
142 8. Differentiable Functions
If c < 0, we get the analogous result with a negative sign. However, f is not differentiable at 0, since
lim h→ 0 +
f (h) − f (0) h
= lim h→ 0 +
h^1 /^2 does not exist.
Example 8.8. The function f : R → R defined by f (x) = x^1 /^3 is differentiable at x 6 = 0 with
f ′(x) =
3 x^2 /^3
To prove this result, we use the identity for the difference of cubes,
a^3 − b^3 = (a − b)(a^2 + ab + b^2 ),
and get for c 6 = 0 that
lim h→ 0
[
f (c + h) − f (c) h
]
= lim h→ 0
(c + h)^1 /^3 − c^1 /^3 h
= lim h→ 0
(c + h) − c h
[
(c + h)^2 /^3 + (c + h)^1 /^3 c^1 /^3 + c^2 /^3
]
= lim h→ 0
(c + h)^2 /^3 + (c + h)^1 /^3 c^1 /^3 + c^2 /^3
=
3 c^2 /^3
However, f is not differentiable at 0, since
lim h→ 0
f (h) − f (0) h
= lim h→ 0
h^2 /^3
does not exist.
Finally, we consider some examples of highly oscillatory functions.
Example 8.9. Define f : R → R by
f (x) =
x sin(1/x) if x 6 = 0, 0 if x = 0.
It follows from the product and chain rules proved below that f is differentiable at x 6 = 0 with derivative
f ′(x) = sin
x
x
cos
x
However, f is not differentiable at 0, since
lim h→ 0
f (h) − f (0) h
= lim h→ 0 sin
h
which does not exist.
Example 8.10. Define f : R → R by
f (x) =
x^2 sin(1/x) if x 6 = 0, 0 if x = 0.
8.1. The derivative 143
−1−1 −0.5 0 0.5 1
−0.
−0.
−0.
−0.
0
1
−0.01−0.1 −0.05 0 0.05 0.
−0.
−0.
−0.
−0.
0
Figure 1. A plot of the function y = x^2 sin(1/x) and a detail near the origin with the parabolas y = ±x^2 shown in red.
Then f is differentiable on R. (See Figure 1.) It follows from the product and chain rules proved below that f is differentiable at x 6 = 0 with derivative
f ′(x) = 2x sin
x
− cos
x
Moreover, f is differentiable at 0 with f ′(0) = 0, since
lim h→ 0
f (h) − f (0) h
= lim h→ 0
h sin
h
In this example, limx→ 0 f ′(x) does not exist, so although f is differentiable on R, its derivative f ′^ is not continuous at 0.
8.1.2. Derivatives as linear approximations. Another way to view Defini- tion 8.1 is to write
f (c + h) = f (c) + f ′(c)h + r(h)
as the sum of a linear (or, strictly speaking, affine) approximation f (c) + f ′(c)h of f (c + h) and a remainder r(h). In general, the remainder also depends on c, but we don’t show this explicitly since we’re regarding c as fixed.
As we prove in the following proposition, the differentiability of f at c is equiv- alent to the condition
lim h→ 0
r(h) h
That is, the remainder r(h) approaches 0 faster than h, so the linear terms in h provide a leading order approximation to f (c + h) when h is small. We also write this condition on the remainder as
r(h) = o(h) as h → 0 ,
pronounced “r is little-oh of h as h → 0.”
Graphically, this condition means that the graph of f near c is close the line through the point (c, f (c)) with slope f ′(c). Analytically, it means that the function
h 7 → f (c + h) − f (c)
8.2. Properties of the derivative 145
8.1.3. Left and right derivatives. For the most part, we will use derivatives that are defined only at the interior points of the domain of a function. Sometimes, however, it is convenient to use one-sided left or right derivatives that are defined at the endpoint of an interval.
Definition 8.14. Suppose that f : [a, b] → R. Then f is right-differentiable at a ≤ c < b with right derivative f ′(c+) if
lim h→ 0 +
[
f (c + h) − f (c) h
]
= f ′(c+)
exists, and f is left-differentiable at a < c ≤ b with left derivative f ′(c−) if
lim h→ 0 −
[
f (c + h) − f (c) h
]
= lim h→ 0 +
[
f (c) − f (c − h) h
]
= f ′(c−).
A function is differentiable at a < c < b if and only if the left and right derivatives at c both exist and are equal.
Example 8.15. If f : [0, 1] → R is defined by f (x) = x^2 , then
f ′(0+) = 0, f ′(1−) = 2.
These left and right derivatives remain the same if f is extended to a function defined on a larger domain, say
f (x) =
x^2 if 0 ≤ x ≤ 1, 1 if x > 1, 1 /x if x < 0.
For this extended function we have f ′(1+) = 0, which is not equal to f ′(1−), and f ′(0−) does not exist, so the extended function is not differentiable at either 0 or
Example 8.16. The absolute value function f (x) = |x| in Example 8.6 is left and right differentiable at 0 with left and right derivatives
f ′(0+) = 1, f ′(0−) = − 1.
These are not equal, and f is not differentiable at 0.
8.2. Properties of the derivative
In this section, we prove some basic properties of differentiable functions.
8.2.1. Differentiability and continuity. First we discuss the relation between differentiability and continuity.
Theorem 8.17. If f : (a, b) → R is differentiable at at c ∈ (a, b), then f is continuous at c.
146 8. Differentiable Functions
Proof. If f is differentiable at c, then
lim h→ 0 f (c + h) − f (c) = lim h→ 0
[
f (c + h) − f (c) h · h
]
= lim h→ 0
[
f (c + h) − f (c) h
]
· lim h→ 0
h
= f ′(c) · 0 = 0,
which implies that f is continuous at c.
For example, the sign function in Example 8.5 has a jump discontinuity at 0 so it cannot be differentiable at 0. The converse does not hold, and a continuous function needn’t be differentiable. The functions in Examples 8.6, 8.8, 8.9 are continuous but not differentiable at 0. Example 9.24 describes a function that is continuous on R but not differentiable anywhere.
In Example 8.10, the function is differentiable on R, but the derivative f ′^ is not continuous at 0. Thus, while a function f has to be continuous to be differentiable, if f is differentiable its derivative f ′^ need not be continuous. This leads to the following definition.
Definition 8.18. A function f : (a, b) → R is continuously differentiable on (a, b), written f ∈ C^1 (a, b), if it is differentiable on (a, b) and f ′^ : (a, b) → R is continuous.
For example, the function f (x) = x^2 with derivative f ′(x) = 2x is continu- ously differentiable on R, whereas the function in Example 8.10 is not continuously differentiable at 0. As this example illustrates, functions that are differentiable but not continuously differentiable may behave in rather pathological ways. On the other hand, the behavior of continuously differentiable functions, whose graphs have continuously varying tangent lines, is more-or-less consistent with what one expects.
8.2.2. Algebraic properties of the derivative. A fundamental property of the derivative is that it is a linear operation. In addition, we have the following product and quotient rules.
Theorem 8.19. If f, g : (a, b) → R are differentiable at c ∈ (a, b) and k ∈ R, then kf , f + g, and f g are differentiable at c with
(kf )′(c) = kf ′(c), (f + g)′(c) = f ′(c) + g′(c), (f g)′(c) = f ′(c)g(c) + f (c)g′(c).
Furthermore, if g(c) 6 = 0, then f /g is differentiable at c with
( f g
(c) =
f ′(c)g(c) − f (c)g′(c) g^2 (c)
148 8. Differentiable Functions
Proof. Since f is differentiable at c, there is a function r(h) such that
f (c + h) = f (c) + f ′(c)h + r(h), lim h→ 0
r(h) h
and since g is differentiable at f (c), there is a function s(k) such that
g (f (c) + k) = g (f (c)) + g′^ (f (c)) k + s(k), lim k→ 0
s(k) k
It follows that
(g ◦ f )(c + h) = g (f (c) + f ′(c)h + r(h)) = g (f (c)) + g′^ (f (c)) · (f ′(c)h + r(h)) + s (f ′(c)h + r(h)) = g (f (c)) + g′^ (f (c)) f ′(c) · h + t(h)
where t(h) = g′^ (f (c)) · r(h) + s (φ(h)) , φ(h) = f ′(c)h + r(h).
Since r(h)/h → 0 as h → 0, we have
lim h→ 0
t(h) h
= lim h→ 0
s (φ(h)) h
We claim that this limit exists and is zero, and then it follows from Proposition 8. that g ◦ f is differentiable at c with
(g ◦ f )′(c) = g′^ (f (c)) f ′(c).
To prove the claim, we use the facts that φ(h) h
→ f ′(c) as h → 0 ,
s(k) k
→ 0 as k → 0.
Roughly speaking, we have φ(h) ∼ f ′(c)h when h is small and therefore
s (φ(h)) h
s (f ′(c)h) h
→ 0 as h → 0.
In detail, let > 0 be given. We want to show that there exists δ > 0 such that ∣ ∣ ∣ ∣
s (φ(h)) h
∣ < ^ if 0^ <^ |h|^ < δ.
First, choose δ 1 > 0 such that ∣ ∣ ∣ ∣
r(h) h
∣ <^ |f^
′(c)| + 1 if 0 < |h| < δ
If 0 < |h| < δ 1 , then
|φ(h)| ≤ |f ′(c)| |h| + |r(h)| < |f ′(c)| |h| + (|f ′(c)| + 1)|h| < (2|f ′(c)| + 1) |h|.
Next, choose η > 0 so that ∣ ∣ ∣ ∣
s(k) k
∣ <^
2 |f ′(c)| + 1
if 0 < |k| < η.
8.3. The chain rule 149
(We include a “1” in the denominator on the right-hand side to avoid a division by zero if f ′(c) = 0.) Finally, define δ 2 > 0 by
δ 2 = η 2 |f ′(c)| + 1
and let δ = min(δ 1 , δ 2 ) > 0.
If 0 < |h| < δ and φ(h) 6 = 0, then 0 < |φ(h)| < η, so
|s (φ(h)) | ≤
|φ(h)| 2 |f ′(c)| + 1
< |h|.
If φ(h) = 0, then s(φ(h)) = 0, so the inequality holds in that case also. This proves that
lim h→ 0
s (φ(h)) h
Example 8.22. Suppose that f is differentiable at c and f (c) 6 = 0. Then g(y) = 1/y is differentiable at f (c), with g′(y) = − 1 /y^2 (see Example 8.4). It follows that the reciprocal function 1/f = g ◦ f is differentiable at c with ( 1 f
(c) = g′(f (c))f ′(c) = −
f ′(c) f (c)^2
The chain rule gives an expression for the derivative of an inverse function. In terms of linear approximations, it states that if f scales lengths at c by a nonzero factor f ′(c), then f −^1 scales lengths at f (c) by the factor 1/f ′(c).
Proposition 8.23. Suppose that f : A → R is a one-to-one function on A ⊂ R with inverse f −^1 : B → R where B = f (A). Assume that f is differentiable at an interior point c ∈ A and f −^1 is differentiable at f (c), where f (c) is an interior point of B. Then f ′(c) 6 = 0 and
(f −^1 )′^ (f (c)) =
f ′(c)
Proof. The definition of the inverse implies that
f −^1 (f (x)) = x.
Since f is differentiable at c and f −^1 is differentiable at f (c), the chain rule implies that (^) ( f −^1
(f (c)) f ′(c) = 1.
Dividing this equation by f ′(c) 6 = 0, we get the result. Moreover, it follows that f −^1 cannot be differentiable at f (c) if f ′(c) = 0.
Alternatively, setting d = f (c), we can write the result as
(f −^1 )′(d) =
f ′^ (f −^1 (d))
Proposition 8.23 is not entirely satisfactory because it assumes the existence and differentiability of an inverse function. We will return to this question in Section 8.7 below, but we end this section with some examples that illustrate the
8.4. Extreme values 151
If f has a (local or global) maximum or minimum at c ∈ A, then f is said to have a (local or global) extreme value at c.
Theorem 7.37 states that a continuous function on a compact set has a global maximum and minimum but does not say how to find them. The following funda- mental result goes back to Fermat.
Theorem 8.27. If f : A ⊂ R → R has a local extreme value at an interior point c ∈ A and f is differentiable at c, then f ′(c) = 0.
Proof. If f has a local maximum at c, then f (x) ≤ f (c) for all x in a δ-neighborhood (c − δ, c + δ) of c, so
f (c + h) − f (c) h
≤ 0 for all 0 < h < δ,
which implies that
f ′(c) = lim h→ 0 +
[
f (c + h) − f (c) h
]
Moreover, f (c + h) − f (c) h ≥ 0 for all −δ < h < 0 ,
which implies that
f ′(c) = lim h→ 0 −
[
f (c + h) − f (c) h
]
It follows that f ′(c) = 0. If f has a local minimum at c, then the signs in these inequalities are reversed, and we also conclude that f ′(c) = 0.
For this result to hold, it is crucial that c is an interior point, since we look at the sign of the difference quotient of f on both sides of c. At an endpoint, we get the following inequality condition on the derivative. (Draw a graph!)
Proposition 8.28. Let f : [a, b] → R. If the right derivative of f exists at a, then: f ′(a+) ≤ 0 if f has a local maximum at a; and f ′(a+) ≥ 0 if f has a local minimum at a. Similarly, if the left derivative of f exists at b, then: f ′(b−) ≥ 0 if f has a local maximum at b; and f ′(b−) ≤ 0 if f has a local minimum at b.
Proof. If the right derivative of f exists at a, and f has a local maximum at a, then there exists δ > 0 such that f (x) ≤ f (a) for a ≤ x < a + δ, so
f ′(a+) = lim h→ 0 +
[
f (a + h) − f (a) h
]
Similarly, if the left derivative of f exists at b, and f has a local maximum at b, then f (x) ≤ f (b) for b − δ < x ≤ b, so f ′(b−) ≥ 0. The signs are reversed for local minima at the endpoints.
In searching for extreme values of a function, it is convenient to introduce the following classification of points in the domain of the function.
152 8. Differentiable Functions
Definition 8.29. Suppose that f : A ⊂ R → R. An interior point c ∈ A such that f is not differentiable at c or f ′(c) = 0 is called a critical point of f. An interior point where f ′(c) = 0 is called a stationary point of f.
Theorem 8.27 limits the search for maxima or minima of a function f on A to the following points.
(1) Boundary points of A. (2) Critical points of f : (a) interior points where f is not differentiable; (b) stationary points where f ′(c) = 0.
Additional tests are required to determine which of these points gives local or global extreme values of f. In particular, a function need not attain an extreme value at a critical point.
Example 8.30. If f : [− 1 , 1] → R is the function
f (x) =
x if − 1 ≤ x ≤ 0, 2 x if 0 < x ≤ 1,
then x = 0 is a critical point since f is not differentiable at 0, but f does not attain a local extreme value at 0. The global maximum and minimum of f are attained at the endpoints x = 1 and x = −1, respectively, and f has no other local extreme values.
Example 8.31. If f : [− 1 , 1] → R is the function f (x) = x^3 , then x = 0 is a critical point since f ′(0) = 0, but f does not attain a local extreme value at 0. The global maximum and minimum of f are attained at the endpoints x = 1 and x = −1, respectively, and f has no other local extreme values.
8.5. The mean value theorem
The mean value theorem is a key result that connects the global behavior of a function f : [a, b] → R, described by the difference f (b) − f (a), to its local behavior, described by the derivative f ′^ : (a, b) → R. We begin by proving a special case.
Theorem 8.32 (Rolle). Suppose that f : [a, b] → R is continuous on the closed, bounded interval [a, b], differentiable on the open interval (a, b), and f (a) = f (b). Then there exists a < c < b such that f ′(c) = 0.
Proof. By the Weierstrass extreme value theorem, Theorem 7.37, f attains its global maximum and minimum values on [a, b]. If these are both attained at the endpoints, then f is constant, and f ′(c) = 0 for every a < c < b. Otherwise, f attains at least one of its global maximum or minimum values at an interior point a < c < b. Theorem 8.27 implies that f ′(c) = 0.
Note that we require continuity on the closed interval [a, b] but differentiability only on the open interval (a, b). This proof is deceptively simple, but the result is not trivial. It relies on the extreme value theorem, which in turn relies on the completeness of R. The theorem would not be true if we restricted attention to functions defined on the rationals Q.
154 8. Differentiable Functions
Proof. If f is increasing and a < x < b, then
f (x + h) − f (x) h
for all sufficiently small h (positive or negative), so
f ′(x) = lim h→ 0
[
f (x + h) − f (x) h
]
Conversely if f ′^ ≥ 0 and a < x < y < b, then by the mean value theorem there exists x < c < y such that
f (y) − f (x) y − x
= f ′(c) ≥ 0 ,
which implies that f (x) ≤ f (y), so f is increasing. Moreover, if f ′(c) > 0, we get f (x) < f (y), so f is strictly increasing.
The results for a decreasing function f follow in a similar way, or we can apply of the previous results to the increasing function −f.
Note that although f ′^ > 0 implies that f is strictly increasing, f is strictly increasing does not imply that f ′^ > 0.
Example 8.37. The function f : R → R defined by f (x) = x^3 is strictly increasing on R, but f ′(0) = 0.
If f is continuously differentiable and f ′(c) > 0, then f ′(x) > 0 for all x in a neighborhood of c and Theorem 8.36 implies that f is strictly increasing near c. This conclusion may fail if f is not continuously differentiable at c.
Example 8.38. Define f : R → R by
f (x) =
x/2 + x^2 sin(1/x) if x 6 = 0, 0 if x = 0.
Then f is differentiable on R with
f ′(x) =
1 / 2 − cos(1/x) + 2x sin(1/x) if x 6 = 0, 1 / 2 if x = 0.
Every neighborhood of 0 includes intervals where f ′^ < 0 or f ′^ > 0, in which f is strictly decreasing or strictly increasing, respectively. Thus, despite the fact that f ′(0) > 0, the function f is not strictly increasing in any neighborhood of 0. As a result, no local inverse of the function f exists on any neighborhood of 0.
8.6. Taylor’s theorem
If f : (a, b) → R is differentiable on (a, b) and f ′^ : (a, b) → R is differentiable, then we define the second derivative f ′′^ : (a, b) → R of f as the derivative of f ′. We define higher-order derivatives similarly. If f has derivatives f (n)^ : (a, b) → R of all orders n ∈ N, then we say that f is infinitely differentiable on (a, b).
Taylor’s theorem gives an approximation for an (n + 1)-times differentiable function in terms of its Taylor polynomial of degree n.
8.6. Taylor’s theorem 155
Definition 8.39. Let f : (a, b) → R and suppose that f has n derivatives
f ′, f ′′,... f (n)^ : (a, b) → R
on (a, b). The Taylor polynomial of degree n of f at a < c < b is
Pn(x) = f (c) + f ′(c)(x − c) +
f ′′(c)(x − c)^2 + · · · +
n!
f (n)(c)(x − c)n.
Equivalently,
Pn(x) =
∑^ n
k=
ak(x − c)k, ak =
k! f (k)(c).
We call ak the kth Taylor coefficient of f at c. The computation of the Taylor polynomials in the following examples are left as an exercise.
Example 8.40. If P (x) is a polynomial of degree n, then Pn(x) = P (x).
Example 8.41. The Taylor polynomial of degree n of ex^ at x = 0 is
Pn(x) = 1 + x +
x^2 · · · +
n!
xn.
Example 8.42. The Taylor polynomial of degree 2n of cos x at x = 0 is
P 2 n(x) = 1 −
x^2 +
x^4 − · · · + (−1)n^
(2n)!
x^2 n.
We also have P 2 n+1 = P 2 n since the Tayor coefficients of odd order are zero.
Example 8.43. The Taylor polynomial of degree 2n + 1 of sin x at x = 0 is
P 2 n+1(x) = x −
x^3 +
x^5 − · · · + (−1)n^
(2n + 1)!
x^2 n+1.
We also have P 2 n+2 = P 2 n+1.
Example 8.44. The Taylor polynomial of degree n of 1/x at x = 1 is
Pn(x) = 1 − (x − 1) + (x − 1)^2 − · · · + (−1)n(x − 1)n.
Example 8.45. The Taylor polynomial of degree n of log x at x = 1 is
Pn(x) = (x − 1) −
(x − 1)^2 +
(x − 1)^3 − · · · + (−1)n+1(x − 1)n.
We write f (x) = Pn(x) + Rn(x).
where Rn is the error, or remainder, between f and its Taylor polynomial Pn. The next theorem is one version of Taylor’s theorem, which gives an expression for the remainder due to Lagrange. It can be regarded as a generalization of the mean value theorem, which corresponds to the case n = 1. The idea of the proof is to subtract a suitable polynomial from the function and apply Rolle’s theorem, just as we proved the mean value theorem by subtracting a suitable linear function.
8.7. * The inverse function theorem 157
which implies that
lim x→ 0
1 − cos x x^2
Note that as well as proving the limit, Taylor’s theorem gives an explicit upper bound for the difference between (1 − cos x)/x^2 and its limit 1/2. For example, ∣ ∣ ∣ ∣
1 − cos(0.1) (0.1)^2
∣ ≤^
Numerically, we have
1 2
1 − cos(0.1) (0.1)^2
In Section 12.7, we derive an alternative expression for the remainder Rn as an integral.
8.7. * The inverse function theorem
The inverse function theorem gives a sufficient condition for a differentiable function f to be locally invertible at a point c with differentiable inverse: namely, that f is continuously differentiable at c and f ′(c) 6 = 0. Example 8.24 shows that one cannot expect the inverse of a differentiable function f to exist locally at c if f ′(c) = 0, while Example 8.38 shows that the condition f ′(c) 6 = 0 is not, on its own, sufficient to imply the existence of a local inverse.
Before stating the theorem, we give a precise definition of local invertibility.
Definition 8.48. A function f : A → R is locally invertible at an interior point c ∈ A if there exist open neighborhoods U of c and V of f (c) such that f |U : U → V is one-to-one and onto, in which case f has a local inverse ( f |U )−^1 : V → U.
The following examples illustrate the definition.
Example 8.49. If f : R → R is the square function f (x) = x^2 , then a local inverse at c = 2 with U = (1, 3) and V = (1, 9) is defined by
( f |U )−^1 (y) =
y.
Similarly, a local inverse at c = −2 with U = (− 3 , −1) and V = (1, 9) is defined by
( f |U )−^1 (y) = −
y.
In defining a local inverse at c, we require that it maps an open neighborhood V of f (c) onto an open neighborhood U of c; that is, we want ( f |U )−^1 (y) to be “close” to c when y is “close” to f (c), not some more distant point that f also maps “close” to f (c). Thus, the one-to-one, onto function g defined by
g : (1, 9) → (− 2 , −1) ∪ [2, 3), g(y) =
y if 1 < y < 4 √ y if 4 ≤ y < 9
is not a local inverse of f at c = 2 in the sense of Definition 8.48, even though g(f (2)) = 2 and both compositions
f ◦ g : (1, 9) → (1, 9), g ◦ f : (− 2 , −1) ∪ [2, 3) → (− 2 , −1) ∪ [2, 3)
are identity maps, since U = (− 2 , −1) ∪ [2, 3) is not a neighborhood of 2.
158 8. Differentiable Functions
Example 8.50. The function f : R → R defined by
f (x) =
cos (1/x) if x 6 = 0 0 if x = 0
is locally invertible at every c ∈ R with c 6 = 0 or c 6 = 1/(nπ) for some n ∈ Z.
Theorem 8.51 (Inverse function). Suppose that f : A ⊂ R → R and c ∈ A is an interior point of A. If f is differentiable in a neighborhood of c, f ′(c) 6 = 0, and f ′^ is continuous at c, then there are open neighborhoods U of c and V of f (c) such that f has a local inverse ( f |U )−^1 : V → U. Furthermore, the local inverse function is differentiable at f (c) with derivative
[( f |U )−^1 ]′^ (f (c)) =
f ′(c)
Proof. Suppose, for definiteness, that f ′(c) > 0 (otherwise, consider −f ). By the continuity of f ′, there exists an open interval U = (a, b) containing c on which f ′^ > 0. It follows from Theorem 8.36 that f is strictly increasing on U. Writing
V = f (U ) = (f (a), f (b)) ,
we see that f |U : U → V is one-to-one and onto, so f has a local inverse on V , which proves the first part of the theorem.
It remains to prove that the local inverse ( f |U )−^1 , which we denote by f −^1 for short, is differentiable. First, since f is differentiable at c, we have
f (c + h) = f (c) + f ′(c)h + r(h)
where the remainder r satisfies
lim h→ 0
r(h) h
Since f ′(c) > 0, there exists δ > 0 such that
|r(h)| ≤
f ′(c)|h| for |h| < δ.
It follows from the differentiability of f that, if |h| < δ,
f ′(c)|h| = |f (c + h) − f (c) − r(h)| ≤ |f (c + h) − f (c)| + |r(h)|
≤ |f (c + h) − f (c)| +
f ′(c)|h|.
Absorbing the term proportional to |h| on the right hand side of this inequality into the left hand side and writing
f (c + h) = f (c) + k,
we find that 1 2
f ′(c)|h| ≤ |k| for |h| < δ.
Choosing δ > 0 small enough that (c − δ, c + δ) ⊂ U , we can express h in terms of k as h = f −^1 (f (c) + k) − f −^1 (f (c)).
