MTH4100 Calculus I
Bill Jackson
School of Mathematical Sciences QMUL
Week 11, Semester 1, 2013
Bill Jackson
Calculus I
Injective (one-to-one) functions
Definition Let f : D → R be a function. Then f is injective (or
one-to-one) if f (x1 ) 6= f (x2 ) whenever x1 6= x2 .
Bill Jackson
Calculus I
Injective (one-to-one) functions
Definition Let f : D → R be a function. Then f is injective (or
one-to-one) if f (x1 ) 6= f (x2 ) whenever x1 6= x2 .
Thus a function is injective if it takes on every value in its range
exactly once.
Bill Jackson
Calculus I
Injective (one-to-one) functions
Definition Let f : D → R be a function. Then f is injective (or
one-to-one) if f (x1 ) 6= f (x2 ) whenever x1 6= x2 .
Thus a function is injective if it takes on every value in its range
exactly once.
Examples: Let f : R → R be defined by f (x) = x 3 and
√
g : R+ → R be defined by g (x) = x, where
R+ = {x ∈ R : x ≥ 0}.
Bill Jackson
Calculus I
Injective (one-to-one) functions
Definition Let f : D → R be a function. Then f is injective (or
one-to-one) if f (x1 ) 6= f (x2 ) whenever x1 6= x2 .
Thus a function is injective if it takes on every value in its range
exactly once.
Examples: Let f : R → R be defined by f (x) = x 3 and
√
g : R+ → R be defined by g (x) = x, where
R+ = {x ∈ R : x ≥ 0}.
The horizontal line test for injective functions: A function is
injective if and only if its graph intersects every horizontal line at
most once. This implies that if f is a continuous function defined
on a closed interval I , then f is injective if and only if f is
monotonic.
Bill Jackson
Calculus I
Injective (one-to-one) functions
Definition Let f : D → R be a function. Then f is injective (or
one-to-one) if f (x1 ) 6= f (x2 ) whenever x1 6= x2 .
Thus a function is injective if it takes on every value in its range
exactly once.
Examples: Let f : R → R be defined by f (x) = x 3 and
√
g : R+ → R be defined by g (x) = x, where
R+ = {x ∈ R : x ≥ 0}.
The horizontal line test for injective functions: A function is
injective if and only if its graph intersects every horizontal line at
most once. This implies that if f is a continuous function defined
on a closed interval I , then f is injective if and only if f is
monotonic.
Examples: Let f : R → R be defined by f (x) = x 2 and
g : [0, π] → R be defined by g (x) = sin x.
Bill Jackson
Calculus I
Injective (one-to-one) functions
Definition Let f : D → R be a function. Then f is injective (or
one-to-one) if f (x1 ) 6= f (x2 ) whenever x1 6= x2 .
Thus a function is injective if it takes on every value in its range
exactly once.
Examples: Let f : R → R be defined by f (x) = x 3 and
√
g : R+ → R be defined by g (x) = x, where
R+ = {x ∈ R : x ≥ 0}.
The horizontal line test for injective functions: A function is
injective if and only if its graph intersects every horizontal line at
most once. This implies that if f is a continuous function defined
on a closed interval I , then f is injective if and only if f is
monotonic.
Examples: Let f : R → R be defined by f (x) = x 2 and
g : [0, π] → R be defined by g (x) = sin x.
Note that if we restrict the domain of f to R+ and the domain of
g to [0, π/2] then the restricted functions will both be injective.
Bill Jackson
Calculus I
Inverse functions
Definition Suppose that f : D → R is an injective function with
range R. Then the inverse function f −1 : R → D is defined by
f −1 (y ) = x whenever f (x) = y .
Note that:
Bill Jackson
Calculus I
Inverse functions
Definition Suppose that f : D → R is an injective function with
range R. Then the inverse function f −1 : R → D is defined by
f −1 (y ) = x whenever f (x) = y .
Note that:
the domain of f −1 is equal to the range of f and the range of
f −1 is equal to the domain of f .
Bill Jackson
Calculus I
Inverse functions
Definition Suppose that f : D → R is an injective function with
range R. Then the inverse function f −1 : R → D is defined by
f −1 (y ) = x whenever f (x) = y .
Note that:
the domain of f −1 is equal to the range of f and the range of
f −1 is equal to the domain of f .
f −1 is read as f inverse.
Bill Jackson
Calculus I
Inverse functions
Definition Suppose that f : D → R is an injective function with
range R. Then the inverse function f −1 : R → D is defined by
f −1 (y ) = x whenever f (x) = y .
Note that:
the domain of f −1 is equal to the range of f and the range of
f −1 is equal to the domain of f .
f −1 is read as f inverse.
f −1 (x) 6= 1/f (x) i.e. f −1 (x) is not the same as f (x)−1 .
Bill Jackson
Calculus I
Inverse functions
Definition Suppose that f : D → R is an injective function with
range R. Then the inverse function f −1 : R → D is defined by
f −1 (y ) = x whenever f (x) = y .
Note that:
the domain of f −1 is equal to the range of f and the range of
f −1 is equal to the domain of f .
f −1 is read as f inverse.
f −1 (x) 6= 1/f (x) i.e. f −1 (x) is not the same as f (x)−1 .
(f −1 ◦ f )(x) = x for all x ∈ D.
Bill Jackson
Calculus I
Inverse functions
Definition Suppose that f : D → R is an injective function with
range R. Then the inverse function f −1 : R → D is defined by
f −1 (y ) = x whenever f (x) = y .
Note that:
the domain of f −1 is equal to the range of f and the range of
f −1 is equal to the domain of f .
f −1 is read as f inverse.
f −1 (x) 6= 1/f (x) i.e. f −1 (x) is not the same as f (x)−1 .
(f −1 ◦ f )(x) = x for all x ∈ D.
(f ◦ f −1 )(y ) = y for all y ∈ R.
Bill Jackson
Calculus I
Method for finding inverse functions
Example: Find the inverse of the function f : R+ → R defined by
f (x) = x 2 .
Bill Jackson
Calculus I
Method for finding inverse functions
Example: Find the inverse of the function f : R+ → R defined by
f (x) = x 2 .
√
Step 1 Solve y = f (x) for x. We have y = x 2 and x ≥ 0 so x = y .
Since the domain and range of f is R+ , we obtain
f −1 : R+ → R+ by x = f −1 (y ) =
Bill Jackson
Calculus I
√
y
Method for finding inverse functions
Example: Find the inverse of the function f : R+ → R defined by
f (x) = x 2 .
√
Step 1 Solve y = f (x) for x. We have y = x 2 and x ≥ 0 so x = y .
Since the domain and range of f is R+ , we obtain
f −1 : R+ → R+ by x = f −1 (y ) =
√
y
Step 2 Relabel x and y so that y is the dependent variable and x is
the independent variable. This gives:
√
f −1 : R+ → R+ by y = f −1 (x) = x
Bill Jackson
Calculus I
Relationship between graphs of f and f −1
Lemma
The graphs of f and f −1 are interchanged by reflection in the line
y = x.
Bill Jackson
Calculus I
Relationship between graphs of f and f −1
Lemma
The graphs of f and f −1 are interchanged by reflection in the line
y = x.
√
Example The graphs of f (x) = x 2 and f −1 (x) = x.
Bill Jackson
Calculus I
Derivatives of inverse functions
Theorem
Suppose that f : D → R is injective, differentiable and f ′ (x) 6= 0
for all x ∈ D. Then f −1 is differentiable and its derivative (f −1 )′
satisfies
1
.
(f −1 )′ (x) = ′ −1
f (f (x))
Equivalently, for all b in the domain of f −1 we have
df −1 1
.
= df −1
dx x=b
dx x=f (b)
Bill Jackson
Calculus I
Derivatives of inverse functions
Theorem
Suppose that f : D → R is injective, differentiable and f ′ (x) 6= 0
for all x ∈ D. Then f −1 is differentiable and its derivative (f −1 )′
satisfies
1
.
(f −1 )′ (x) = ′ −1
f (f (x))
Equivalently, for all b in the domain of f −1 we have
df −1 1
.
= df −1
dx x=b
dx x=f (b)
Example f : R+ → R by f (x) = x 2 .
Bill Jackson
Calculus I
The Natural Logarithm Function
Definition Consider the function f (x) = x −1 . This is continuous
on the closed interval [a, b] for any 0 < a < b.
Bill Jackson
Calculus I
The Natural Logarithm Function
Definition Consider the function f (x) = x −1 . This is continuous
on the closed interval [a, b] for any 0 < a < b.
The Fundamental
Theorem of Calculus (Part 1) now tells us that
R x −1
F (x) = 1 t dt is continuous on [a, b] and differentiable on (a, b)
for all 0 < a < b.
Bill Jackson
Calculus I
The Natural Logarithm Function
Definition Consider the function f (x) = x −1 . This is continuous
on the closed interval [a, b] for any 0 < a < b.
The Fundamental
Theorem of Calculus (Part 1) now tells us that
R x −1
F (x) = 1 t dt is continuous on [a, b] and differentiable on (a, b)
for all 0 < a < b.
This function F is an important function: it is called the natural
logarithm function and is denoted by ln. Thus
Z x
ln x =
t −1 dt .
1
Bill Jackson
Calculus I
Properties of the natural logarithm function
Lemma
The domain of ln x is (0, ∞), its derivative is x −1 , and it is an
increasing function on (0, ∞).
Bill Jackson
Calculus I
Rules for manipulating natural logarithms
Lemma
Suppose a, x are positive real numbers. Then
1
2
3
4
ln ax = ln a + ln x.
ln x1 = − ln x.
ln xa = ln a − ln x.
ln x q = q ln x for any rational number q.
Bill Jackson
Calculus I
Rules for manipulating natural logarithms
Lemma
Suppose a, x are positive real numbers. Then
1
2
3
4
ln ax = ln a + ln x.
ln x1 = − ln x.
ln xa = ln a − ln x.
ln x q = q ln x for any rational number q.
Examples:
1
2
3
4
ln 8 + ln cos x =
z2 + 3
=
ln
2z − 1
ln cot x =
√
ln 5 x − 3 =
Bill Jackson
Calculus I
Range of the natural logarithm function
Lemma
The range of ln x is (−∞, ∞).
Bill Jackson
Calculus I
Range of the natural logarithm function
Lemma
The range of ln x is (−∞, ∞).
Definition The fact that the range of ln x is (−∞, ∞) implies in
particular that ln x = 1 for some x ∈ (0, ∞). The point e for which
ln e = 1 is referred to as Euler’s constant or the base of the natural
logarithm. Its approximate numerical value is
e = 2.718281828459 . . .
Bill Jackson
Calculus I
Antiderivatives involving the natural logarithm
We have seen that ln x is an antiderivative for 1/x for any interval
I ⊂ (0, ∞). Our next result extends this to all intervals which do
not contain zero.
Bill Jackson
Calculus I
Antiderivatives involving the natural logarithm
We have seen that ln x is an antiderivative for 1/x for any interval
I ⊂ (0, ∞). Our next result extends this to all intervals which do
not contain zero.
Theorem
Let I be an interval. If 0 6∈ I then ln |x| is an antiderivative for
f (x) = 1/x on I . More generally , if g (x) is non-zero and
differentiable on I , then ln |g (x)| is an antiderivative for
g ′ (x)/g (x) on I .
Bill Jackson
Calculus I
Antiderivatives involving the natural logarithm
We have seen that ln x is an antiderivative for 1/x for any interval
I ⊂ (0, ∞). Our next result extends this to all intervals which do
not contain zero.
Theorem
Let I be an interval. If 0 6∈ I then ln |x| is an antiderivative for
f (x) = 1/x on I . More generally , if g (x) is non-zero and
differentiable on I , then ln |g (x)| is an antiderivative for
g ′ (x)/g (x) on I .
Examples
(a) Suppose I is an interval such that cos x 6= 0 for all x ∈ I . Then
ln | sec x| is an antiderivative for tan x on I .
(b)
Z 1
1
dx =
x
−
3
0
Bill Jackson
Calculus I
The Exponential Function
Definition The natural logarithm function is injective and hence is
invertible. Its inverse function exp(x) = ln−1 (x) is called the
exponential function.
Bill Jackson
Calculus I
Properties of the exponential function
Lemma
The domain of exp x is R and its range is (0, ∞). It is continuous
d
exp x = exp x.
and differentiable on R and satisfies dx
Bill Jackson
Calculus I
Irrational powers of real numbers
Lemma
Suppose a is a positive real number and q ∈ Q. Then
aq = exp(q ln a) .
Bill Jackson
Calculus I
Irrational powers of real numbers
Lemma
Suppose a is a positive real number and q ∈ Q. Then
aq = exp(q ln a) .
Definition For any a ∈ R with a > 0, the exponential function
with base a, ax , is defined by putting
ax = exp(x ln a)
for all x ∈ R.
Bill Jackson
Calculus I
Irrational powers of real numbers
Lemma
Suppose a is a positive real number and q ∈ Q. Then
aq = exp(q ln a) .
Definition For any a ∈ R with a > 0, the exponential function
with base a, ax , is defined by putting
ax = exp(x ln a)
for all x ∈ R.
This definition implies that
ln(ax ) = ln[exp(x ln a)] = x ln a
for all x ∈ R.
Bill Jackson
Calculus I
Properties of the exponential function with base a
Lemma
Suppose a, b, c ∈ R and a > 0. Then:
1
2
ab · ac = ab+c :
(ab )c = abc .
Bill Jackson
Calculus I
Properties of the exponential function with base a
Lemma
Suppose a, b, c ∈ R and a > 0. Then:
1
2
ab · ac = ab+c :
(ab )c = abc .
Theorem
Suppose that a > 0 and a 6= 1. Then the exponential function with
base a is differentiable for all x ∈ R and satisfies
d x
a = ax ln a .
dx
Hence
Z
ax dx =
Bill Jackson
ax
+C.
ln a
Calculus I
Logarithms to the base a
d x
Definition When a > 1, dx
a = ax ln a > 0 and hence f (x) = ax
is strictly increasing for all x ∈ R.
Bill Jackson
Calculus I
Logarithms to the base a
d x
Definition When a > 1, dx
a = ax ln a > 0 and hence f (x) = ax
is strictly increasing for all x ∈ R.
When 0 < a < 1, a similar argument shows that f (x) = ax is
strictly decreasing for all x ∈ R.
Bill Jackson
Calculus I
Logarithms to the base a
d x
Definition When a > 1, dx
a = ax ln a > 0 and hence f (x) = ax
is strictly increasing for all x ∈ R.
When 0 < a < 1, a similar argument shows that f (x) = ax is
strictly decreasing for all x ∈ R.
This implies that f (x) = ax is injective for any fixed a > 0 with
a 6= 1. Hence its inverse function exists. This inverse function is
called the logarithm of x to the base a and is denoted by loga x.
Bill Jackson
Calculus I
Logarithms to the base a
d x
Definition When a > 1, dx
a = ax ln a > 0 and hence f (x) = ax
is strictly increasing for all x ∈ R.
When 0 < a < 1, a similar argument shows that f (x) = ax is
strictly decreasing for all x ∈ R.
This implies that f (x) = ax is injective for any fixed a > 0 with
a 6= 1. Hence its inverse function exists. This inverse function is
called the logarithm of x to the base a and is denoted by loga x.
We have
loga (ax ) = x = aloga x
for all x ∈ R, by the definition of an inverse function.
Bill Jackson
Calculus I
Logarithms to the base a
d x
Definition When a > 1, dx
a = ax ln a > 0 and hence f (x) = ax
is strictly increasing for all x ∈ R.
When 0 < a < 1, a similar argument shows that f (x) = ax is
strictly decreasing for all x ∈ R.
This implies that f (x) = ax is injective for any fixed a > 0 with
a 6= 1. Hence its inverse function exists. This inverse function is
called the logarithm of x to the base a and is denoted by loga x.
We have
loga (ax ) = x = aloga x
for all x ∈ R, by the definition of an inverse function.
This gives
ln x = ln aloga x = loga x · ln a .
and hence
loga x =
Bill Jackson
ln x
ln a
Calculus I
Further properties of the exponential function
The definition of ax gives us an alternative notation for exp(x).
Recall that 1 = ln e where e is Euler’s constant. This implies that
e x = exp(x ln e) = exp x.
We often use e x for exp x.
Bill Jackson
Calculus I
Further properties of the exponential function
The definition of ax gives us an alternative notation for exp(x).
Recall that 1 = ln e where e is Euler’s constant. This implies that
e x = exp(x ln e) = exp x.
We often use e x for exp x.
We have seen that
d x
dx e
= e x . This gives
Bill Jackson
Calculus I
R
e x dx = e x + C .
Further properties of the exponential function
The definition of ax gives us an alternative notation for exp(x).
Recall that 1 = ln e where e is Euler’s constant. This implies that
e x = exp(x ln e) = exp x.
We often use e x for exp x.
R
d x
We have seen that dx
e = e x . This gives e x dx = e x + C .
We can now use the chain rule to deduce:
Lemma
Let f (x) be a differentiable function. Then
d f (x)
e
= e f (x) f ′ (x)
dx
and
Z
e f (x) f ′ (x)dx = e f (x) + C .
Bill Jackson
Calculus I
The number e as a limit
Theorem
e = lim (1 + x)1/x
x→0
Bill Jackson
Calculus I
Read
Thomas’ Calculus:
Section 7.7 Inverse trigonometric functions,
and Section 7.8, Hyperbolic functions
You will need this information for
coursework 10!
Bill Jackson
Calculus I