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Consequences of Differentiability, the Mean Value Theorem

Module by: Lawrence Baggett. E-mail the author

Summary: The first derivative test for extreme values is proven, followed by the mean value theorem, the inverse function theorem, and some exercises pertaining to these theorems and proofs.

Definition 1:

Let f:S→Rf:S→R be a real-valued function of a real variable, and let cc be an element of the interior of S.S. Then ff is said to attain a local maximum at cc if there exists a δ>0δ>0 such that (c-δ,c+δ)⊆S(c-δ,c+δ)⊆S and f(c)≥f(x)f(c)≥f(x) for all x∈(c-δ,c+δ).x∈(c-δ,c+δ).

The function ff is said to attain a local minimum at cc if there exists an interval (c-δ,c+δ)⊆S(c-δ,c+δ)⊆S such that f(c)≤f(x)f(c)≤f(x) for all x∈(c-δ,c+δ).x∈(c-δ,c+δ).

The next theorem should be a familiar result from calculus.

Theorem 1: First Derivative Test for Extreme Values

Let f:S→Rf:S→R be a real-valued function of a real variable, and let c∈Sc∈S be a point at which ff attains a local maximum or a local minimum. If ff is differentiable at c,c, then f'(c)f'(c) must be 0.0.

Proof

We prove the theorem when ff attains a local maximum at c.c. The proof for the case when ff attains a local minimum is completely analogous.

Thus, let δ>0δ>0 be such that f(c)≥f(x)f(c)≥f(x) for all xx such that |x-c|<δ.|x-c|<δ. Note that, if nn is sufficiently large, then both c+1nc+1n and c-1nc-1n belong to the interval (c-δ,c+δ).(c-δ,c+δ). We evaluate f'(c)f'(c) in two ways. First,

f ' ( c ) = lim n f ( c + 1 n ) - f ( c ) 1 n ≤ 0 f ' ( c ) = lim n f ( c + 1 n ) - f ( c ) 1 n ≤ 0
(1)

because the numerator is always nonpositive and the denominator is always positive. On the other hand,

f ' ( c ) = lim n f ( c - 1 n ) - f ( c ) - 1 n ≥ 0 f ' ( c ) = lim n f ( c - 1 n ) - f ( c ) - 1 n ≥ 0
(2)

since both numerator and denominator are nonpositive. Therefore, f'(c)f'(c) must be 0, as desired.

Of course we do not need a result like Theorem 1 for functions of a complex variable, since the derivative of every real-valued function of a complex variable necessarily is 0, independent of whether or not the function attains an extreme value.

REMARK As mentioned earlier, the zeroes of a function are often important numbers. The preceding theorem shows that the zeroes of the derivative f'f' of a function ff are intimately related to finding the extreme values of the function f.f. The zeroes of f'f' are often called the critical points for f.f. Part (a) of the Exercise 1 establishes the familiar procedure from calculus for determining the maximum and minimum of a continuous real-valued function on a closed interval.

Exercise 1

  1. Let ff be a continuous real-valued function on a closed interval [a,b],[a,b], and assume that ff is differentiable at each point xx in the open interval (a,b).(a,b). Let MM be the maximum value of ff on this interval, and mm be its minimum value on this interval. Write SS for the set of all x∈(a,b)x∈(a,b) for which f'(x)=0.f'(x)=0. Suppose xx is a point of [a,b][a,b] for which f(x)f(x) is either MM or m.m. Prove that xx either is an element of the set S,S, or xx is one of the endpoints aa or b.b.
  2. Let ff be the function defined on [0,1/2)[0,1/2) by f(t)=t/(1-t).f(t)=t/(1-t). Show that f(t)<1f(t)<1 for all t∈[0,1/2).t∈[0,1/2).
  3. Let t∈(-1/2,1/2)t∈(-1/2,1/2) be given. Prove that there exists an r<1,r<1, depending on t,t, such that |t/(1+y)|<r|t/(1+y)|<r for all yy between 0 and t.t.
  4. Let tt be a fixed number for which 0<t<1.0<t<1. Show that, for all 0≤s≤t,0≤s≤t,(t-s)/(1+s)≤t.(t-s)/(1+s)≤t.

Probably the most powerful theorem about differentiation is the next one. It is stated as an equation, but its power is usually as an inequality; i.e., the absolute value of the left hand side is less than or equal to the absolute value of the right hand side.

Theorem 2: Mean Value Theorem

Let ff be a real-valued continuous function on a closed bounded interval [a,b],[a,b], and assume that ff is differentiable at each point xx in the open interval (a,b).(a,b). Then there exists a point c∈(a,b)c∈(a,b) such that

f ( b ) - f ( a ) = f ' ( c ) ( b - a ) . f ( b ) - f ( a ) = f ' ( c ) ( b - a ) .
(3)

Proof

This proof is tricky. Define a function hh on [a,b][a,b] by

h ( x ) = x ( f ( b ) - f ( a ) ) - f ( x ) ( b - a ) . h ( x ) = x ( f ( b ) - f ( a ) ) - f ( x ) ( b - a ) .
(4)

Clearly, hh is continuous on [a,b][a,b] and is differentiable at each point x∈(a,b).x∈(a,b). Indeed,

h ' ( x ) = f ( b ) - f ( a ) - f ' ( x ) ( b - a ) . h ' ( x ) = f ( b ) - f ( a ) - f ' ( x ) ( b - a ) .
(5)

It follows from this equation that the theorem will be proved if we can show that there exists a point c∈(a,b)c∈(a,b) for which h'(c)=0.h'(c)=0. Note also that

h ( a ) = a ( f ( b ) - f ( a ) ) - f ( a ) ( b - a ) = a f ( b ) - b f ( a ) h ( a ) = a ( f ( b ) - f ( a ) ) - f ( a ) ( b - a ) = a f ( b ) - b f ( a )
(6)

and

h ( b ) = b ( f ( b ) - f ( a ) ) - f ( b ) ( b - a ) = a f ( b ) - b f ( a ) , h ( b ) = b ( f ( b ) - f ( a ) ) - f ( b ) ( b - a ) = a f ( b ) - b f ( a ) ,
(7)

showing that h(a)=h(b).h(a)=h(b).

Let mm be the minimum value attained by the continuous function hh on the compact interval [a,b][a,b] and let MM be the maximum value attained by hh on [a,b].[a,b]. If m=M,m=M, then hh is a constant on [a,b][a,b] and h'(c)=0h'(c)=0 for all c∈(a,b).c∈(a,b). Hence, the theorem is true if M=m,M=m, and we could use any c∈(a,b).c∈(a,b). If m≠M,m≠M, then at least one of these two extreme values is not equal to h(a).h(a). Suppose m≠h(a).m≠h(a). Of course, mm is also not equal to h(b).h(b). Let c∈[a,b]c∈[a,b] be such that h(c)=m.h(c)=m. Then, in fact, c∈(a,b).c∈(a,b). By Theorem 1, h'(c)=0.h'(c)=0.

We have then that in every case there exists a point c∈(a,b)c∈(a,b) for which h'(c)=0.h'(c)=0. This completes the proof.

REMARK The Mean Value Theorem is a theorem about real-valued functions of a real variable, and we will see later that it fails for complex-valued functions of a complex variable. (See part (f) of (Reference).) In fact, it can fail for a complex-valued function of a real variable. Indeed, if f(x)=u(x)+iv(x)f(x)=u(x)+iv(x) is a continuous complex-valued function on the interval [a,b],[a,b], and differentiable on the open interval (a,b),(a,b), then the Mean Value Theorem certainly holds for the two real-valued functions uu and v,v, so that we would have

f ( b ) - f ( a ) = u ( b ) - u ( a ) + i ( v ( b ) - v ( a ) ) = u ' ( c 1 ) ( b - a ) + i v ' ( c 2 ) ( b - a ) , f ( b ) - f ( a ) = u ( b ) - u ( a ) + i ( v ( b ) - v ( a ) ) = u ' ( c 1 ) ( b - a ) + i v ' ( c 2 ) ( b - a ) ,
(8)

which is not f'(c)(b-a)f'(c)(b-a) unless we can be sure that the two points c1c1 and c2c2 can be chosen to be equal. This simply is not always possible. Look at the function f(x)=x2+ix3f(x)=x2+ix3 on the interval [0,1].[0,1].

On the other hand, if ff is a real-valued function of a complex variable (two real variables), then a generalized version of the Mean Value Theorem does hold. See part (c) of (Reference).

One of the first applications of the Mean Value Theorem is to show that a function whose derivative is identically 0 is necessarily a constant function. This seemingly obvious fact is just not obvious. The next exercise shows that this result holds for complex-valued functions of a complex variable, even though the Mean Value Theorem does not.

Exercise 2

  1. Suppose ff is a continuous real-valued function on (a,b)(a,b) and that f'(x)=0f'(x)=0 for all x∈(a,b).x∈(a,b). Prove that ff is a constant function on (a,b).(a,b). HINT: Show that f(x)=f(a)f(x)=f(a) for all x∈[a,b]x∈[a,b] by using the Mean Value Theorem applied to the interval [a,x].[a,x].
  2. Let ff be a complex-valued function of a real variable. Suppose ff is differentiable at each point xx in an open interval (a,b),(a,b), and assume that f'(x)=0f'(x)=0 for all x∈(a,b).x∈(a,b). Prove that ff is a constant function. HINT: Use the real and imaginary parts of f.f.
  3. Let ff be a complex-valued function of a complex variable, and suppose that ff is differentiable on a disk Br(c)⊆C,Br(c)⊆C, and that f'(z)=0f'(z)=0 for all z∈Br(c).z∈Br(c). Prove that f(z)f(z) is constant on Br(c).Br(c). HINT: Let zz be an arbitrary point in Br(c),Br(c), and define a function h:[0,1]→Ch:[0,1]→C by h(t)=f((1-t)c+tz).h(t)=f((1-t)c+tz). Apply part (b) to h.h.

The next exercise establishes, at last, two important identities.

Exercise 3

(cos2+sin2=1cos2+sin2=1 and exp(iπ=-1.exp(iπ=-1.)

  1. Prove that cos2(z)+sin2(z)=1cos2(z)+sin2(z)=1 for all complex numbers z.z.
  2. Prove that cos(π)=-1.cos(π)=-1. HINT: We know from part (a) that cos(π)=±1.cos(π)=±1. Using the Mean Value Theorem for the cosine function on the interval [0,π],[0,π], derive a contradiction from the assumption that cos(π)=1.cos(π)=1.
  3. Prove that exp(iπ)=-1.exp(iπ)=-1. HINT: Recall that exp(iz)=cos(z)+isin(z)exp(iz)=cos(z)+isin(z) for all complex z.z. (Note that this does not yet tell us that eiπ=-1.eiπ=-1. We do not yet know that exp(z)=ez.exp(z)=ez.)
  4. Prove that cosh2z-sinh2z=1cosh2z-sinh2z=1 for all complex numbers z.z.
  5. Compute the derivatives of the tangent and hyperbolic tangent functions tan=sin/costan=sin/cos and tanh=sinh/cosh.tanh=sinh/cosh. Show in fact that
    tan'=1cos2andtanh'=1cosh2.tan'=1cos2andtanh'=1cosh2.
    (9)

Here are two more elementary consequences of the Mean Value Theorem.

Exercise 4

  1. Suppose ff and gg are two complex-valued functions of a real (or complex) variable, and suppose that f'(x)=g'(x)f'(x)=g'(x) for all x∈(a,b)x∈(a,b) (or x∈Br(c).x∈Br(c).) Prove that there exists a constant kk such that f(x)=g(x)+kf(x)=g(x)+k for all x∈(a,b)x∈(a,b) (or x∈Br(c).x∈Br(c).)
  2. Suppose f'(z)=cexp(az)f'(z)=cexp(az) for all z,z, where cc and aa are complex constants with a≠0.a≠0. Prove that there exists a constant c'c' such that f(z)=caexp(az)+c'.f(z)=caexp(az)+c'. What if a=0?a=0?
  3. (A generalization of part (a)) Suppose ff and gg are continuous real-valued functions on the closed interval [a,b],[a,b], and suppose there exists a partition {x0<x1<...<xn}{x0<x1<...<xn} of [a,b][a,b] such that both ff and gg are differentiable on each subinterval (xi-1,xi).(xi-1,xi). (That is, we do not assume that ff and gg are differentiable at the endpoints.) Suppose that f'(x)=g'(x)f'(x)=g'(x) for every xx in each open subinterval (xi-1,xi).(xi-1,xi). Prove that there exists a constant kk such that f(x)=g(x)+kf(x)=g(x)+k for all x∈[a,b].x∈[a,b]. HINT: Use part (a) to conclude that f=g+hf=g+h where hh is a step function, and then observe that hh must be continuous and hence a constant.
  4. Suppose ff is a differentiable real-valued function on (a,b)(a,b) and assume that f'(x)≠0f'(x)≠0 for all x∈(a,b).x∈(a,b). Prove that ff is 1-1 on (a,b).(a,b).

Exercise 5

Let f:[a,b]→Rf:[a,b]→R be a function that is continuous on its domain [a,b][a,b] and differentiable on (a,b).(a,b). (We do not suppose that f'f' is continuous on (a,b).(a,b).)

  1. Prove that ff is nondecreasing on [a,b][a,b] if and only if f'(x)≥0f'(x)≥0 for all x∈(a,b).x∈(a,b). Show also that ff is nonincreasing on [a,b][a,b] if and only if f'(x)≤0f'(x)≤0 for all x∈(a,b).x∈(a,b).
  2. Conclude that, if f'f' takes on both positive and negative values on (a,b),(a,b), then ff is not 1-1. (See the proof of (Reference).)
  3. Show that, if f'f' takes on both positive and negative values on (a,b),(a,b), then there must exist a point c∈(a,b)c∈(a,b) for which f'(c)=0.f'(c)=0. (If f'f' were continuous, this would follow from the Intermediate Value Theorem. But, we are not assuming here that f'f' is continuous.)
  4. Prove the Intermediate Value Theorem for Derivatives: Suppose ff is continuous on the closed bounded interval [a,b][a,b] and differentiable on the open interval (a,b).(a,b). If f'f' attains two distinct values v1=f'(x1)<v2=f'(x2),v1=f'(x1)<v2=f'(x2), then f'f' attains every value vv between v1v1 and v2.v2. HINT: Suppose vv is a value between v1v1 and v2.v2. Define a function gg on [a,b][a,b] by g(x)=f(x)-vx.g(x)=f(x)-vx. Now apply part (c) to g.g.

Here is another perfectly reasonable and expected theorem, but one whose proof is tough.

Theorem 3: Inverse Function Theorem

Suppose f:(a,b)→Rf:(a,b)→R is a function that is continuous and 1-1 from (a,b)(a,b) onto the interval (a',b').(a',b'). Assume that ff is differentiable at a point c∈(a,b)c∈(a,b) and that f'(c)≠0.f'(c)≠0. Then f-1f-1 is differentiable at the point f(c),f(c), and

f - 1 ' ( f ( c ) ) = 1 f ' ( c ) . f - 1 ' ( f ( c ) ) = 1 f ' ( c ) .
(10)

Proof

The formula f-1'(f(c))=1/f'(c)f-1'(f(c))=1/f'(c) is no surprise. This follows directly from the Chain Rule. For, if f-1(f(x))=x,f-1(f(x))=x, and ff and f-1f-1 are both differentiable, then f-1'(f(c))f'(c)=1,f-1'(f(c))f'(c)=1, which gives the formula. The difficulty with this theorem is in proving that the inverse function f-1f-1 of ff is differentiable at f(c).f(c). In fact, the first thing to check is that the point f(c)f(c) belongs to the interior of the domain of f-1,f-1, for that is essential if f-1f-1 is to be differentiable there, and here is where the hypothesis that ff is a real-valued function of a real variable is important. According to (Reference), the 1-1 continuous function ff maps [a,b][a,b] onto an interval [a',b'],[a',b'], and f(c)f(c) is in the open interval (a',b'),(a',b'), i.e., is in the interior of the domain of f-1.f-1.

According to part (2) of (Reference), we can prove that f-1f-1 is differentiable at f(c)f(c) by showing that

lim x → f ( c ) f - 1 ( x ) - f - 1 ( f ( c ) ) x - f ( c ) = 1 f ' ( c ) . lim x → f ( c ) f - 1 ( x ) - f - 1 ( f ( c ) ) x - f ( c ) = 1 f ' ( c ) .
(11)

That is, we need to show that, given an ϵ>0,ϵ>0, there exists a δ>0δ>0 such that if 0<|x-f(c)|<δ0<|x-f(c)|<δ then

| f - 1 ( x ) - f - 1 ( f ( c ) ) x - f ( c ) - 1 f ' ( c ) | < ϵ . | f - 1 ( x ) - f - 1 ( f ( c ) ) x - f ( c ) - 1 f ' ( c ) | < ϵ .
(12)

First of all, because the function 1/q1/q is continuous at the point f'(c),f'(c), there exists an ϵ'>0ϵ'>0 such that if |q-f'(c)|<ϵ',|q-f'(c)|<ϵ', then

| 1 q - 1 f ' ( c ) | < ϵ | 1 q - 1 f ' ( c ) | < ϵ
(13)

Next, because ff is differentiable at c,c, there exists a δ'>0δ'>0 such that if 0<|y-c|<δ'0<|y-c|<δ' then

| f ( y ) - f ( c ) y - c - f ' ( c ) | < ϵ ' . | f ( y ) - f ( c ) y - c - f ' ( c ) | < ϵ ' .
(14)

Now, by (Reference), f-1f-1 is continuous at the point f(c),f(c), and therefore there exists a δ>0δ>0 such that if |x-f(c)|<δ|x-f(c)|<δ then

| f - 1 ( x ) - f - 1 ( f ( c ) | < δ ' | f - 1 ( x ) - f - 1 ( f ( c ) | < δ '
(15)

So, if |x-f(c)|<δ,|x-f(c)|<δ, then

| f - 1 ( x ) - c | = | f - 1 ( x ) - f - 1 ( f ( c ) ) | < δ ' . | f - 1 ( x ) - c | = | f - 1 ( x ) - f - 1 ( f ( c ) ) | < δ ' .
(16)

But then, by Equation 14,

| f ( f - 1 ( x ) ) - f ( c ) f - 1 ( x ) - c - f ' ( c ) | < ϵ ' , | f ( f - 1 ( x ) ) - f ( c ) f - 1 ( x ) - c - f ' ( c ) | < ϵ ' ,
(17)

from which it follows, using Equation 13, that

| f - 1 ( x ) - f - 1 ( f ( c ) ) x - f ( c ) - 1 f ' ( c ) | < ϵ , | f - 1 ( x ) - f - 1 ( f ( c ) ) x - f ( c ) - 1 f ' ( c ) | < ϵ ,
(18)

as desired.

REMARK A result very like Theorem 3 is actually true for complex-valued functions of a complex variable. We will have to show that if cc is in the interior of the domain SS of a one-to-one, continuously differentiable, complex-valued function ff of a complex variable, then f(c)f(c) is in the interior of the domain f(S)f(S) of f-1.f-1. But, in the complex variable case, this requires a somewhat more difficult argument. Once that fact is established, the proof that f-1f-1 is differentiable at f(c)f(c) will be the same for complex-valued functions of complex variables as it is here for real-valued functions of a real variable. Though the proof of Theorem 3 is reasonably complicated for real-valued functions of a real variable, the corresponding result for complex functions is much more deep, and that proof will have to be postponed to a later chapter. See (Reference).

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