Lecture 15:
Last time: MVT.
Special case: Rolle’s Theorem (when f (a) = f (b)).
Recall:
Defn: Let f be defined on an interval I.
f is increasing (or strictly increasing) if whenever x1, x2 ∈ I and
x2 > x1, then f (x2) > f (x1).
– Note: I could be open, closed, half-open, half-closed, infinite, R.
Another application of MVT:
Theorem 12, p. 140:
If f is cts. on [a, b] and diffble. on (a, b) and f 0(x) > 0 for all
x ∈ (a, b), then f is increasing on [a, b].
Recall Proof: Let a ≤ x1 < x2 ≤ b. By MVT, there exists c s.t.
x1 < c < x2 and
f (x2) − f (x1)
= f 0(c) > 0.
x2 − x1
Since x2 − x1 > 0, we have f (x2) − f (x1) > 0. Note: in above, we applied MVT to a = x1, b = x2.
Theorem 12 is stated more generally in text: can replace [a, b] by
any interval J := (a, b) ∪ S where S ⊆ {a, b}. Require f to be
diffble. on (a, b) and cts. on J.
In particular, if f is diffble. on an open interval I and f 0(x) > 0
for all x ∈ I then f is increasing on I.
Q: If f is differentiable and increasing on an open interval I, is
f 0(x) > 0 for all x in the interval? No.
1
f (x) = x3: (problem 21, p. 143):
f 0(x) = 3x2 and so f 0(0) = 0.
Claim: f is increasing on R.
Proof: Since f 0(x) > 0 on (0, ∞) it is increasing on (0, ∞). Similarly it is increasing on (−∞, 0). Finally, if x2 > 0 > x1, then
x32 > 0 > x31. Another proof:
x32 − x31 = (x2 − x1)(x22 + x1x2 + x21)
Suffices to show: for all x1, x2 s.t. (x1, x2) 6= (0, 0),
x22 + x1x2 + x21 > 0.
HW5.
Text defines:
decreasing (strictly decreasing): if whenever x1, x2 ∈ I and x2 >
x1, then f (x1) > f (x2).
non-decreasing: if whenever x1, x2 ∈ I and x2 > x1, then
f (x2) ≥ f (x1).
non-increasing: if whenever x1, x2 ∈ I and x2 > x1, then f (x2) ≤
f (x1).
and the proof of Theorem 12 carries over to analogues of these.
if f 0(x) < 0, then decreasing;
if f 0(x) ≥ 0, then non-decreasing;
if f 0(x) ≤ 0, then non-increasing;
Examples: of a non-decreasing diffble function :
2
x x≥0
f (x) =
0 x<0
2
(diffble. because f+(0) = 0 = f−(0)).
As pointed out by a student:
Proposition: If f is non-decreasing and differentiable on an open
interval I, then f 0(x) ≥ 0 for all x ∈ I.
Proof: For x ∈ I,
f (x + h) − f (x)
f 0(x) = lim
≥0
h
h→0+
Since for sufficiently small h > 0, x + h ∈ I and since f is nondecreasing, f (x + h) − f (x) ≥ 0.
Implicit Differentiation
A curve in the xy-plane is usually described as the set of all points
(x, y) that satisfy a given equation F (x, y) = 0.
Examples: 1. y − x2 = 0,
2. x2 + y 2 − 4 = 0.
Often the equation F (x, y) = 0 determines y implicitly as a function y = f (x), ie., F (x, f (x)) = 0. The function f is found by
solving for y in terms of x. Sometimes there is more than one function f that works.
√
Examples: 1.√ y = f (x) = x2.
2. y = f1(x) = 4 − x2,
y = f2(x) = − 4 − x2.
In order for this to work so that the “implicit function” f is differentiable, the two-variable function F must satisfy a two-variable
differentiability property that we will not delve into.
In implicit differentiation,, goal is to find f 0(x) without explicitly
finding f (x).
Namely, differentiate both sides of F (x, y) = 0 with respect to x
dy
in terms of x and y.
and then solve for y 0 = dx
This should give y 0 for all functions f determined implicitly by
F (x, y) = 0.
3
Example 1: y − x2 = 0.
y 0 − 2x = 0 and so y 0 = 2x.
Example 2: x2 + y 2 − 4 = 0.
2x + 2yy 0 = 0 and so y 0 = − xy .
Here, we have used the chain rule:
y 2 = (f (x))2 for some unknown function f ; by the chain rule
(y 2)0 = ((f (x))2)0 = 2f (x)f 0(x) = 2yy 0
Verify in terms of derivatives of explicit functions:
√
y = f1(x) = 4 − x2:
x
1
=−
y
4 − x2
√
and the same holds for y = f2(x) = − 4 − x2.
y 0 = (1/2)(−2x) √
Note that the implicit function may not be differentiable when
y = 0, which is indeed the case for f1 and f2.
4
Lecture 16:
In some examples, the curve is defined by G(x, y) = H(x, y),
which is equivalent to F ((x, y) = 0, where F (x, y) = G(x, y) −
H(x, y):
Example 3: x sin(xy − y) = x2 − 1.
sin(xy − y) + x(cos(xy − y))(y + xy 0 − y 0) = 2x
y 0(x − 1)x(cos(xy − y)) = 2x − sin(xy − y) − x(cos(xy − y))(y)
2x − sin(xy − y) − x(cos(xy − y))(y)
y0 =
(x − 1)x(cos(xy − y))
Higher Order Implicit Differentiation:
Differentiate again:
Example 2: x2 + y 2 − 4 = 0. Recall that y 0 =
0
−y + xy
y =
=
y2
00
−y + x −x
y
y2
2
=
−y − xy
y2
−x
y .
So,
x2 + y 2
4
=−
=
−
y3
y3
Power rule for rational exponents: recall that we proved the power
rule only for integer exponents (although we have been using it for
rational exponents).
Proposition: Let m and n be integers with n 6= 0. Let f (x) =
xm/n. Then f 0(x) = (m/n)x(m/n)−1.
The function y = f (x) = xm/n is an implicit function of y n −xm =
0. Then
ny n−1y 0 = mxm−1
So
mxm−1 m xm−1
y =
=
ny n−1
n y n−1
0
5
m xm−1
m m−1+(m(1−n)/n)
=
=
x
n (xm/n)n−1
n
m
= x(m/n)−1
n
Recall that we will later define and prove the power rule for f (x) =
xr for all real r. So far, we have shown this only for rational r.
One-to-One functions and increasing/decreasing functions:
Proposition: Let I be an interval. Let f be increasing on I or
decreasing I. Then f is one-to-one on I.
Proof:
Case 1: f is increasing. If x1 6= x2, then one is larger than the
other. WLOG x2 > x1.
Then f (x2) > f (x1) and so f (x2) 6= f (x1).
Thus, f is one-to-one.
Case 2: f is decreasing. Then g(x) := −f (x) is increasing and so
g is one-to-one.
Then f is one-to-one: if x1 6= x2, then g(x2) 6= g(x1) and so
f (x2) 6= f (x1). The kettle.
Q: If f is one-to-one on an interval I, must it be increasing on I
or decreasing I?
A: No
Example:
f (x) =
x 0≤x≤1
−x 1 < x ≤ 2
f is one-to-one because its graph passes the horizontal line test.
6
But f is neither increasing nor decreasing.
Proposition: If f is one-to-one and continuous on an interval I, it
must be increasing on I or decreasing on I.
Outline of Proof:
Special case: I is a closed interval: I = [a, b].
Since f is one-to-one, f (a) 6= f (b).
Case 1: f (b) > f (a). Show that f is increasing on I.
Proof by contradiction: if f is not increasing, then there exist c, d
s.t. a ≤ c < d ≤ b and f (c) > f (d).
By intermediate value theorem, f is not one-to-one on [a, b].
Case 2: f (b) < f (a). Show that f is decreasing on I.
Apply Case 1 to g(x) := −f (x). Then −f is increasing and so f
is decreasing.
Can extend proof from closed intervals to arbitrary intervals: f is
increasing on an interval iff it is increasing on all closed sub-intervals.
7
Lecture 17:
Inverse functions
Defn. Let f be one-to-one on an interval I. We define the inverse
function f −1 by f −1(y) is the (unique) x in the domain of f s.t.
y = f (x).
x = f −1(y) iff y = f (x)
i.e., you “solve” for x in terms of y from the equation y = f (x)
Examples: 1. f (x) = 3x is 1-1 on R. So, the inverse function
is obtained by solving for x in terms of y in the eqn. y = 3x; so
x = (1/3)y is the inverse function.
2. f (x) = x2 is 1-1 on [0, ∞). So, the inverse function is obtained
√
by solving for x in terms of y in the eqn. y = x2; so x = y is the
inverse function.
3. f (x) = tan(x) is 1-1 on (−π/2, π/2): since f 0(x) = sec2(x) > 0
on (−π/2, π/2), by Theorem 12 f is increasing on (−π/2, π/2), and
so is 1-1 on (−π/2, π/2). The inverse function is x = arctan(y).
Usually, we write use the letter x for the domain variable and y
for the range variable.
So, if f (x) = 3x on R then f −1(y) = (1/3)y, but we write
f −1(x) = (1/3)x, i.e., we find the inverse function x = f −1(y) and
then reverse the roles of x and y and write y = f −1(x).
√
Similarly, if f (x) = x2 on [0, ∞), we write f −1(x) = x.
And if f (x) = tan(x) on (π/2, π/2), we write f −1(x) = arctan x.
Note that the domain of f −1 is the range of f and the range of
f −1 is the domain of f .
In terms of graphs, the graph of f −1 is obtained from the graph
of f by reflecting across the line y = x, i.e., interchanging the x and
8
y coordinates:
Graph of f = {(x, f (x)) : x ∈ Dom(f )} = {(f −1(y), y) : y ∈ Ran(f )}
Reflect across x = y and we get:
{(y, f −1(y)) : y ∈ Ran(f )} = {(x, f −1(x)) : x ∈ Dom(f −1)}
= Graph of f −1
√
Draw graphs of 3x, (1/3)x, x2, x, tan(x), arctan(x).
Caution: Do not confuse f −1(x) with
1
f (x) .
Properties:
(f −1)−1 = f .
f −1(f (x)) = x = f (f −1(x))
If f and f −1 are differentiable, then
1 = f 0(f −1(x))((f −1)0(x))
So
1
(f −1)0(x) =
f 0(f −1(x))
√
Example: f (x) = x2 and f −1(x) = x.
(f −1)0(x) =
1
f 0(f −1(x))
=
(1)
1
1
√
=
2f −1(x) 2 x
But this assumes that f −1 is diffble at x and f 0(f −1(x)) 6= 0.
Inverse function theorem: If f is diffble. on an open interval I and
either f 0(x) > 0 for all x ∈ I or f 0(x) < 0 for all x ∈ I, then f −1 is
diffble. on the range of f |I and (??) holds.
Proof is a bit involved.
Another example: Let f (x) = x3 + x. Find (f −1)0(10).
9
((f −1)0(x) =
1
f 0(f −1(x))
=
1
3(f −1(x))2 + 1
(f −1)(10) = 2
(f −1)0(10) = 1/13
Recall f (x) = xr can be defined for all real r.
Similarly, g(x) = ax can be defined for all a > 0.
Consider the difference between f (x) = x2 and g(x) = 2x.
f (10) = 100, g(10) ≈ 1000, f (20) = 400, g(20) ≈ 1, 000, 000.
f (x) is a polynomial and g(x) is an exponential function. Exponential functions grow much faster than polynomials.
Example: f (x) = 2x; then f −1(x) = log2(x):
y = 2x ⇔ x = log2(y); the switch the roles of x and y.
Draw graphs.
Brief review of laws of exponents and laws of logarithms and limits
at infinity for exponential functions and logarithmic functions.
For a > 1,
lim ax = ∞, lim ax = 0
x→∞
x→−∞
For 0 < a < 1,
lim ax = 0, lim ax = ∞
x→∞
x→−∞
For a > 1,
lim loga(x) = ∞, lim loga(x) = −∞
x→∞
x→0+
For 0 < a < 1,
lim loga(x) = −∞, lim loga(x) = ∞
x→0+
x→∞
10
Lecture 18:
Will not cover antiderivatives and integrals until later.
Defn: For x > 0, let Ax be the area of the region in the yt-plane
bounded by the curve y = 1/t, the t-axis, and the vertical lines t = 1
and t = x. Define
Ax
x≥1
ln(x) =
−Ax 0 < x < 1
Note that the domain of ln is (0, ∞), f (x) > 0 for x > 1, f (x) < 0
for 0 < x < 1, and f (1) = 0.
Note that these properties all hold for f (x) = loga(x) when a > 1.
Theorem 1 (p. 175): Let f (x) = ln(x). Then
1
x
This is a special case of the Fundamental Theorem of Calculus
(chapter 5).
f 0(x) =
Proof: Let x > 0. Apply the definition:
ln(x + h) − ln(x)
d
ln(x) = lim
h→0
dx
h
From the picture, we see that for h > 0,
h
h
< ln(x + h) − ln(x) < ;
x+h
x
this clear if x > 1, but in fact it is true for all x. Thus,
1
ln(x + h) − ln(x)
1
<
<
x+h
h
x
By the squeeze theorem, f+0 (x) = x1 .
11
A similar picture shows that for h < 0 and x + h > 0,
h
h
> ln(x + h) − ln(x) >
x
x+h
So,
1
ln(x + h) − ln(x)
1
<
<
x
h
x+h
Thus, f−0 (x) = x1 . Thus, f 0(x) = x1 . Corollary: ln(x) is an increasing function.
Theorem 2: Properties of ln(x):
1. ln(xy) = ln(x) + ln(y)
2. ln(1/x) = − ln(x).
3. ln(x/y) = ln(x) − ln(y)
4. ln(xr ) = r ln(x) (for now, only for rational r)
Deja vu from loga(x)?
Lemma (Theorem 13, p .141, an application of MVT): If f is
continuous on an interval I and f 0(x) = 0 for all interior points x of
I, then f is constant on I.
Proof: Let x0, x1 ∈ I s.t. x0 < x1. By MVT, there exists c s.t.
x0 < c < x1 and
f (x1) − f (x0)
= f 0(c) = 0.
x1 − x0
So, f (x1) = f (x0). So, f is constant. Proof of Theorem 2:
(i) Fix y > 0, and let g(y) = ln(xy) − ln(x).
By the Chain Rule,
d
y
1
g 0(y) = (ln(xy) − ln(x)) =
− = 0.
dx
xy x
12
By the Lemma, ln(xy) − ln(x) is a constant (as a function of x).
But for x = 1, we have ln(xy) − ln(x) = ln(y) − ln(1) = ln(y).
Thus, for all x > 0, ln(xy) − ln(x) = ln(y).
(ii) By (i), ln(x) + ln(1/x) = ln(x(1/x)) = ln(1) = 0.
(iii) By (i), ln(y)+ln(x/y) = ln(x). Thus, ln(x/y) = ln(x)−ln(y).
(iv) By (i), this holds for all positive integers r. It then follows from
(ii) that this holds for all negative integers r. And ln(x0) = ln(1),
which we have already noted is 0. So, (iv) holds for all integers.
For a rational m/n, we have (xm/n)n = xm. Thus, since (iv) holds
for integers, we have
n ln(xm/n) = ln((xm/n)n) = ln(xm) = m ln(x).
Thus, ln(xm/n) = (m/n) ln(x). Corollary: 1) limx→∞ ln(x) = ∞ and 2) limx→0+ ln(x) = −∞
Proof:
1): For a positive integer n, log(2n) = n log(2) can be made arbitrarily large (by choosing n large). If x > 2n, then, since ln(x) is
increasing, ln(x) > n log(2) can be made arbitrarily large.
Rigorous proof: Let N > 0. Let K be any integer ≥ N/ ln(2).
Let M := 2K .
If x > M , then since ln(x) is increasing,
ln(x) > ln(M ) = K ln(2) ≥ (N/ ln(2)) ln(2) = N.
2): As x → 0+, 1/x → ∞ and so ln(x) = − ln(1/x) → −∞. 13