Math 341 Lecture Notes on Chapter 5 The Derivative. 5.2: Derivative and the Intermediate Value Property

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Math 341 – Lecture Notes on Chapter 5 – The Derivative

Lecture 26

§5.2: Derivative and the Intermediate Value Property

Definition of the Derivative

Let g: A → R be a function defined on an interval A. Given c ∈ A, the

derivative of g at c is defined by g0(c) = lim

x→c

g(x)−g(c)

x−c ,

provided this limit exists.

An interpretation of g0(c) as a tangent line to the curve y = g(x) is depicted in the figure below:

Note: Alternative Definitions of g0(c) In Exercise 5.2.6, you verify that

g0(c) = lim t→c g(t)−g(c) t−c = limh→0 g(c+h)−g(c) h = limh→0 g(c+h)−g(c−h) 2h . 1

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Lipschitz Functions

Let A be an interval. A function f : A → R is said to be Lipschitz if there exists a positive constant K such that

|f(x)−f(y)| < K|x−y| for all x, y ∈ A.

This implies that the difference quotients are bounded: f(x)−f(y)

x−y < K,

which comes close to differentiability. In fact, a Lipschitz function (although not necessarily differentiable) is uniformly continuous on A. If 0 < c <1 and

|g(x)−g(y)| < c|x−y| for all x, y ∈ R,

then g is called a contraction mapping. In this case (see Exercise 4.3.11), there will exist some x0 ∈ A such that g(x0) = x0. The point x0 is called a

fixed point of g.

Examples

(a) f(x) =xn where n∈ N. Recall that

xn−cn = (x−c) xn−1 +c1xn−2 +c2xn−3 +· · ·+cn−2x1 +cn−1. So, f0(c) = lim x→c xn−cn x−c = limx→c x n−1+c1xn−2+c2xn−3+· · ·+cn−2x1+cn−1 = ncn−1 (b) f(x) =|x|. Then f0(c) =        1 if c > 0, −1 if c < 0, DNE if c = 0.

So, continuity at a point does NOT imply differentiability.

Theorem (differentiability implies continuity)

If g: A → R is differentiable at c ∈ A, then g is continuous at c.

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Proof. By hypothesis,

f0(c) = lim

x→c

g(x)−g(c) x−c

exits. Also, we know that limx→c(x−c) = 0. By the Algebraic Limit Theorem,

lim x→c f(x)−f(c) = lim x→c f(x)−f(c) x−c ·(x−c) ALT = lim x→c g(x)−g(c) x−c ·limx→c(x−c) = f0(c)·0 = 0.

So, limx→cf(x) =f(c), which implies that f is continuous at c. Algebraic Differentiability Theorem

Let f and g be functions defined on an interval A. Assume f and g are differ-entiable at c ∈ A. Then 1. (f +g)0(c) =f0(c) +g0(c). 2. (kf)0(c) =kf0(c) for all k ∈ R. 3. (Product Rule) (f g)0(c) = f0(c)g(c) +f(c)g0(c). 4. (Quotient Rule) f g 0 (c) = f 0(c)g(c)f(c)g0(c) [g(c)]2 .

Proof. The proofs are all fairly basic. As an illustration we’ll only go through the proof of the product rule.

(f g)(x)−(f g)(c) x−c = f(x)g(x)−f(c)g(c) x−c = f(x)g(x)−f(c)g(x) +f(c)g(x)−f(c)g(c) x−c = f(x)−f(c) x−c g(x) +f(c)g(x)−g(c) x−c

Taking the limit as x →c and using the continuity of g give

(f g)0(c) =f0(c)g(c) +f(c)g0(c).

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Chain Rule

Let f: A → R and g: B → R satisfy f(A) ⊆ B so that g ◦f : A → R. If f is differentiable at c and g is differentiable at f(c), then g ◦f is differentiable at

c and

(g ◦f)0(c) =g0(f(c))f0(c).

Proof. Since g is differentiable at f(c), g0(f(c)) = lim

y→f(c)

g(y)−g(f(c)) y −f(c) . Define the function d: B →R by

d(y) = (g(y)−g(f(c)) y−f(c) if y ∈ B and y 6= f(c), g0(f(c)) if y = f(c). (∗) Then lim y→f(c) d(y) = g0(f(c))

and so d is continuous at f(c). Now rewrite equation (∗) as

g(y)−g(f(c)) =d(y)[y −f(c)].

The last equation is true for all y. Substitute f(t) for y and divide by t −c for t 6= c to get g(f(t))−g(f(c)) t−c = d(f(t)) f(t)−f(c) t−c (t6= c).

Apply the Algebraic Limit Theorem to take the limit as t →c to get

(g ◦f)0(c) =g0(f(c))f0(c).

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Lecture 27

§5.2: Derivative and the Intermediate Value Property

(Con-tinued)

Let’s look at another proof that differentiability implies continuity. This time we’ll use the -δ definition directly without using the Algebraic Limit Theorem.

Theorem (Differentiability Implies Continuity)

Let g: A→ R be differentiable at c ∈ A, where A is an interval. Then g is also continuous at c.

Proof. Let > 0 be given. We wish to show that there exists δ > 0 such that

f(x)−f(c) < whenever |x−c| < δ and x ∈ A. Since g0(c) = limx→c f(x)−f(c)

x−c , there exists α > 0 such that if 0 < |x −c| < α and

x ∈ A, then f(x)−f(c) x−c −f 0(c) < , which implies f0(c)− < f(x)−f(c) x−c < f 0(c) + and so f(x)−f(c) ≤ |f0(c)|+ |x−c|. (∗) Set δ = min α, |f0(c)|+

. Then if |x−c| < δ and x ∈ A, we have

f(x)−f(c) ≤ |f0(c)|+ |x−c| by inequality (∗) since |x−c| < δ ≤ α < |f0(c)|+ |f0(c)|+ since |x−c| < δ ≤ |f0(c)|+ ≤.

This proves that g is continuous at c.

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Example (Derivative exists everywhere but is not continuous every-where)

Consider the function

f(x) =

(

x2sin(1/x) if x 6= 0,

0 if x = 0.

If x 6= 0,

f0(x) = 2xsin(1/x) +x2cos(1/x)(−1/x2) = 2xsin(1/x)−cos(1/x).

If x = 0, f0(0) = lim t→0 f(t)−f(0) t−0 = limt→0 t2sin(1/t)−0 t−0 = limt→0tsin(1/t) = 0. Thus, f is everywhere differentiable and

f0(x) =

(

2xsin(1/x)−cos(1/x) if x 6= 0,

0 if x = 0.

Note that f0(x) exists at all x ∈ R, however f0(x) is not everywhere continuous since limx→0f0(x) does not exist, butf0(0) does exists. Sof0(x) is discontinuous at x = 0. To verify the nonexistence of the limit, we can use the two sequences

(xn) = 1 2nπ → 0 f0(xn) =−1, (yn) = 1 (2n+ 1)π →0 f0(yn) = 1.

Graphs, created with Mathematica, of the functions in this example are included on the following two pages.

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Derivative Example

f

(

x

) =

x

2

sin

(

1

/

x

)

.

In[1]:= f[x_]:=x2*Sin1

x

For large x, the curve y=x2sin(1/x) asymptotically approaches the line y=x.

In[2]:= Plot[{f[x], x},{x,-2, 2}] Out[2]= -2 -1 1 2 -2 -1 1 2

For small x, the curve y=x2sin(1/x) oscillates between the curves y= -x2 and y=x2.

In[3]:= Plotx2*Sin1

x , x2,-x2,x, -1 4 , 1 4  Out[3]= -0.2 -0.1 0.1 0.2 -0.06 -0.04 -0.02 0.02 0.04 0.06

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In[4]:= D[f[x], x] Out[4]= -Cos1 x +2 x Sin1 x 

For large x, the derivative curve y=f'(x) asymptotically approaches the line y=1.

In[5]:= Plot-Cos1

x

+2 x Sin1

x

, 1,{x,-5, 5}, PlotRange→All, PlotPoints→200

Out[5]= -4 -2 2 4 -1.0 -0.5 0.5 1.0

For small x, the curve y=f'(x) oscillates approximately between the lines y= -1 and y=1.

In[6]:= Plot-Cos1

x

+2 x Sin1

x

,-1, 1,{x,-1, 1}, PlotRange→All, PlotPoints→200

Out[6]= -1.0 -0.5 0.5 1.0 -1.0 -0.5 0.5 1.0

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Interior Extremum Theorem

Suppose f is differentiable on an open interval (a, b) and attains a maximum (or a minimum) at c ∈ (a, b). Then f0(c) = 0.

Proof. Suppose f attains a maximum at c ∈ (a, b). Let (xn) and (yn) be sequence

in (a, b) with

a < xn < c < yn < b

and limxn = limyn = c. Then

f0(c) = lim n→∞ f(xn)−f(c) xn −c ≥ 0 since f(xn)−f(c) ≤ 0 and xn−c < 0, and f0(c) = lim n→∞ f(yn)−f(c) yn −c ≤ 0 since f(yn)−f(c) ≤0 and yn −c > 0. So, f0(c) = 0.

The proof in the case when f attains a minimum at c ∈ (a, b) is entirely similar.

Next we will see that there is an intermediate value property for derivatives.

Darboux’s Theorem

Assume f is differentiable on a closed interval [a, b] and f0(a) < α < f0(b) [or f0(a) > α > f0(b)], then there exists c ∈ (a, b) such that f0(c) = α.

Note: As we saw in an earlier example, the derivative can exist everywhere on [a, b] yet not be continuous everywhere on [a, b]. So, the proof cannot assume that f0 is continuous. However, the proof will use both the differentiability of f and the continuity of f.

Proof. Assume f0(a) < α < f0(b). Consider the new function g(x) := f(x)−αx.

Then

g0(x) =f0(x)−α

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and

g0(a) < 0< g0(b). For the positive number 0 =

|g0(a)|

2 , there exists a δ > 0 (with 0 < δ < b−a) such that if a < x < a+δ, then g(x)−g(a) x−a < g 0(a) + 0 = g0(a) + |g0(a)| 2 = g0(a) 2 < 0, which implies g(x)−g(a) < g 0(a) 2 x−a) < 0. Thus there exists at least one point x ∈ (a, b) where

g(a) > g(x).

By a similar argument, there exists at least one y ∈ (a, b) such that

g(y) < g(b).

The continuous function g: [a, b] →R has a minimum by the Extreme Value The-orem, but this minimum is not at x = a or x = b. Thus there exists c ∈ (a, b) such that g has a minimum at c. But then

g0(c) = f0(c)−α = 0

So f0(c) = α.

The proof in the case f0(a) > α > f0(b) is entirely similar.

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Lecture 28

§5.3 Mean Value Theorems

The most important version of the Mean Value Theorem is the following:

Theorem (Mean Value Theorem)

If f: [a, b] → R is continuous on [a, b] and differentiable on (a, b), then there exists a point c ∈ (a, b) where

f0(c) = f(b)−f(a)

b−a .

Here is a picture depicting the theorem:

To prove the Mean Value Theorem, we need a lemma:

Lemma (Rolle’s Theorem)

Let f: [a, b] → R be continuous on [a, b] and differentiable on (a, b). If f(a) = f(b), then there exists a point c ∈ (a, b) where f0(c) = 0.

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Proof. Since f is continuous on [a, b], which is compact, f attains a maximum and a minumum on [a, b]. If the maximum and minimum occur at the endpoints, then f is a constant and f0(x) = 0 for all x ∈ (a, b). In this case we may choose any c ∈ (a, b) and then f0(c) = 0.

If either the maximum or minimum occurs at an interior point c ∈ (a, b), then by the Interior Extremum Theorem, f0(c) = 0.

Now we’ll prove the Mean Value Theorem. The following figure will be helpful for understanding the proof.

Proof of Mean Value Theorem. The equation of the line through the points (a, f(a)) and (b, f(b)) is y = f(b)−f(a) b−a (x−a) +f(a). 12

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Consider the difference between this line and the curve y = f(x). Set d(x) := f(x)− f(b)−f(a) b−a (x−a) +f(a) .

Then d is continuous on [a, b], differentiable on (a, b), and d(a) = d(b). By Rolle’s Theorem, there exists c ∈ (a, b) where d0(c) = 0. Since

d0(x) = f0(x)− f(b)−f(a) b−a , we get 0 = f0(c)− f(b)−f(a) b−a . Corollary

If g: A → R is differentiable on an interval A and satisfies g0(x) = 0 for all x ∈ A, then g(x) = k for some constant k ∈ R.

Proof. Let x, y ∈ A where x < y. By the Mean Value Theorem applied to g on the interval [x, y], there exists c ∈ (x, y) such that

0 = g0(c) = f(y)−f(x)

y−x

which implies g(x) = g(y). Set k to this common value. Since x and y were arbitrary, g(x) =k for all x ∈ A.

Corollary

Suppose f and g are differentiable functions on an interval A and f0(x) = g0(x) for all x ∈ A. Then f(x) =g(x) +k for some constant k.

Proof. Let h(x) = f(x) −g(x). Then h0(x) = f0(x)−g0(x) = 0 for all x ∈ A. By the previous corollary, h(x) = k for some constant k. So, f(x) = g(x) +k for some constant k ∈ R and all x ∈ A.

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Lecture 29

First, recall the Mean Value Theorem that we studied last time:

Theorem (Mean Value Theorem)

If f: [a, b] → R is continuous on [a, b] and differentiable on (a, b), then there exists a point c ∈ (a, b) where

f0(c) = f(b)−f(a)

b−a .

Let’s study an example.

Example

Recall that a function f: A → R is Lipschitz on A if there exists an M > 0

such that f(x)−f(y) x−y ≤ M

for all x, y ∈ A with x 6= y.

Assume f is twice-differentiable on [a, b]. Thus f0 is differentiable on [a, b] and so f0 is also continuous on the compact set [a, b]. Hence f0 is bounded on [a, b] and so there exists some M > 0 such that |f0(x)| ≤ M for all x ∈ [a, b].

Let x, y ∈ [a, b] where x < y. By the Mean Value Theorem, there exists c ∈ (a, b) such that f(x)−f(y) x−y = f 0(c). Then f(x)−f(y) x−y = |f0(c)| ≤ M.

Since x and y were arbitrary, f is Lipschitz on [a, b]. Thus, we have shown that twice-differentiable functions defined on a finite closed interval are Lipschitz on that interval.

Now consider a pair of function f: A →R and g: A → Rthat are both continuous on [a, b] and differentiable on (a, b). By the Mean Value Theorem, there exist

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c1, c2 ∈ (a, b) such that f0(c1) = f(b)−f(a) b−a and g 0 (c2) = g(b)−g(a) b−a .

If g(a) 6= g(b), we can divide to obtain f0(c1) g0(c2)

= f(b)−f(a) g(b)−g(a).

In fact, a stronger statement is true. It turns out to be possible to choose c1 = c2 in the last equation (but not in the earlier equations).

Generalized Mean Value Theorem

If f and g are continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c ∈ (a, b) such that

f(b)−f(a)g0(c) = g(b)−g(a)f0(c).

If g0 is never zero on (a, b), this can be stated as f0(c)

g0(c) =

f(b)−f(a) g(b)−g(a).

Proof. Consider the function

h(x) = [f(b)−f(a)]g(x)−[g(b)−g(a)]f(x).

Then h is continuous on [a, b] and differentiable on (a, b) and h(a) = f(b)g(a)−f(a)g(b) = h(b).

By Rolle’s Theorem, there exists c ∈ (a, b) such that

0 =h0(c) = [f(b)−f(a)]g0(c)−[g(b)−g(a)]f0(c)

and so

[f(b)−f(a)]g0(c) = [g(b)−g(a)]f0(c).

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L’Hospital’s Rule : 0/0 Case

Let f and g be continuous on an interval containing a. Assume f and g are differentiable on this initerval with the possible exception of the point a. If f(a) = g(a) = 0 and g0(x) 6= 0 for all x 6= a, then

lim x→a f0(x) g0(x) = L implies limxa f(x) g(x) = L.

Proof. (The proof of this version of L’Hospital’s book is an exercise in the textbook.) Assume

lim

x→a

f0(x) g0(x) = L.

For every > 0 there exists δ such that if 0 < |x−a| < δ, then

f0(x) g0(x) −L < .

Now for any particular x satisfying 0 < |x−a| < δ, by the Generalized Mean Value Theorem, there exists c between a and x such that

f0(c) g0(c) = f(x)−f(a) g(x)−g(a) = f(x) g(x). But then f(x) g(x) −L = f0(c) g0(c) −L < (since 0 < |c−a| < |x−a| < δ.) So, lim x→a f(x) g(x) = L. Examples lim x→0 sinx x = limx→0 cosx 1 = 1 1. lim x→0 x−sinx x3 = limx→0 1−cosx 3x2 = limx→0 sinx 6x = limx→0 cosx 6 = 1 6.

We’ll state the next L’Hospital Rule as a one-sided limit.

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L’Hospital’s Rule: ∞/∞ Case

Assume f and g are differntiable on (a, b) and that g0(x) 6= 0 for all x ∈ (a, b). If limx→ag(x) = ∞ (or −∞), then

lim x→a f0(x) g0(x) = L implies xlim→a f(x) g(x) = L.

Proof. The proof is given in the textbook.

Exercise 5.3.6

(a) Let g: [0, a] → R be differentiable, g(0) = 0, and |g0(x)| ≤ M for all x ∈ [0, a]. Show that |g(x)| ≤M x for all x ∈ [0, a].

(b) Let h: [0, a] → R be twice differentiable, h0(0) = h(0) = 0, and |h00(x)| ≤

M for all x ∈ [0, a]. Show that |h(x)| ≤ M x2 2 for all x ∈ [0, a].

(c) Let f: [0, a] →R be three times differentiable,

0 = f(0) = f0(0) = f00(0) and |f000(x)| ≤ M for all x ∈ [0, a].

Show that |f(x)| ≤ M x3

3! for all x ∈ [0, a].

Proof of (a). Let x ∈ (0, a]. By the Mean Value Theorem there exists c1 ∈ (0, x) such that g0(c1) = g(x)−g(0) x−0 = g(x) x . Then g(x) =xg0(c1) |g(x)| = M x since |g0(c1)| ≤ M.

By the continuity of g at 0, the inequality holds for all x∈ [0, a].

Proof of (b). Let x ∈ (0, a]. We apply the Generalized Mean Value Theorem to the pair of functions h(x) and x2. There exists c1 ∈ (0, x) such that

h0(c1) 2c1 = h(x)−h(0) x2 02 = h(x) x2 ⇒ h(x) = h0(c1)x2 2c1 . 17

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By the Mean Value Theorem there exists c2 ∈ (0, c1) such that h00(c2) = h 0(c 1)−h0(0) c1 −0 = h 0(c 1) c1 ⇒ h0(c1) = c1h00(c2).

Combining these results gives h(x) = h 0(c 1)x2 2c1 = c1h 00(c 2)x2 2c1 = h 00(c 2)x2 2 |h(x)| ≤ M x 2 2 for x ∈ (0, a].

By the continuity of h at 0, the inequality holds for all x ∈ [0, a].

Proof of (c). Let x ∈ (0, a]. We apply the Generalized Mean Value Theorem to the pair of functions f(x) and x3. There exists c1 ∈ (0, x) such that

f0(c1) 3c21 = f(x)−f(0) x3 03 ⇒ f(x) = f0(c1)x3 3c21 .

Again apply the Generalized Mean Value Theorem to the pair of functions f0(x) and x2. There exists c2 ∈ (0, c1) such that

f00(c2) 2c2 = f 0(c1)f0(0) c21 −02 = f0(c1) c21 ⇒ f 0(c 1) = f00(c2)c21 2c2 .

Finally, apply the Mean Value Theorem to the function f00 on the interval (0, c2). There exists c3 ∈ (0, c2) such that

f000(c3) = f00(c2)−f00(0) c2 −0 = f 00(c2) c2 ⇒ f00(c2) = c2f000(c3).

Putting the pieces together gives f(x) = f0(c1)· x3 3c21 = f 00(c 2)· c21 2c2 · x 3 3c21 = f 000(c 2)c2 · c21 2c2 · x 3 3c21 = f 000(c 3)· x3 3!. Since |f000(c3)| ≤ M, this gives

|f(x)| ≤ M x

3 3!

for x ∈ (0, a]. By continuity at 0, the inequality holds for all x ∈ [0, a].

Figure

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