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3
2
Definition. The function f is continuous at x = a if lim f ( x ) = f ( a ).
This definition is equivalent to the condition lim f ( x ) = lim f ( x ) = f ( a ).
x → a
Intuitively, continuous functions have a graph which can be drawn without
lifting the pencil.
x → a —^ x → a +
Chapter 4.
CONTINUOUS FUNCTIONS
4.1 Continuity
Example 4.1. Define f
x = 1.
( 1 ) so that the function f
x
2 1 ( x ) = x^3 — 1
is continuous at
Solution. By the definition, f ( x ) will be continuous at x = 1 if lim f ( x ) = f ( 1 ).
lim f^ ( x )^ =^ lim^
x — 1 = lim x^ +^1 = 2.
x → 1
x → 1 x → 1 x^3 — 1 x → 1 x^2 + x + 1 3
Therefore, f ( x ) will be continuous at x = 1 if f ( 1 ) =
2 .
Define f ( 0 ) so that the following functions will be continuous at x = 0. √ 1 + x — 1
1 + 2 x — 1
4.1. f ( x ) = x
. 4.2. f ( x ) = √ 3 1 + 2 x — 1
4.3. f ( x ) =
sin x
. 4.4. f ( x ) = 4
1 — cos x .
x x^2
4.5. f ( x ) =
tan 2 x
. 4.6. f ( x ) = sin x sin
x x
4.7. f ( x ) = ( 1 + x )^1 /x^ ( x > 0 ). 4.8. f ( x ) = e —^1 /x^2.
4.9. If the function f ( x ) is continuous for all x and f ( x ) =
x
2 — 5 x + 4 x — 4
when
x /= 4 , what is f ( 4 )?
4.10. If possible, define f
uous at x = 1.
( 1 ) so that the function f ( x ) =
x^2 — 2 x + 1
x^2 — 4 x + 3
is contin-
Example 4.2. Determine the value of k such that the function
f ( x ) =
3 kx — 5 , x < 2;
4 x — 5 k, x ≥ 2
22 CHAPTER 4. CONTINUOUS FUNCTIONS
11
is continuous.
Solution. Consider the point x = 2.
lim x → 2 —
lim x → 2 +
f ( x ) = lim x → 2 —
f ( x ) = lim x → 2 +
( 3 kx — 5 ) = 6 k — 5;
( 4 x — 5 k ) = 8 — 5 k ;
f ( 2 ) = 4 × 2 — 5 k = 8 — 5 k.
The function f ( x ) will be continuous at x = 2 if lim x → 2 —^
f ( x ) = lim x → 2 +^
f ( x ) = f ( 2 ),
so for continuity put
6 k — 5 = 8 — 5 k ⇒ 11 k = 13 ⇒ k =
13 .
Find the values of all unknown constants so that the function is continuous.
4.11. f x
x + 1 , x ≤ 1 ,
3 — mx
2 , x > 1_._
4.12. f ( x ) =
4 kx — 4 , x > 2 ,
4 x — 2 k, x ≤ 2_._
4.13. f ( x ) =
x
2 9
x — 3
, x /= 3 , (^) 4.14. f ( x ) =
e^2 x, x^ <^^0 ,
4.15. f ( x ) =
A, x = 3_._
x
2
bx ln x, x > 2_._
4.16. f ( x ) =
x — a, x ≥ 0_._
e
2 x + d , x ≥ 0 ,
x + 2 , x < 0_._
24 CHAPTER 4. CONTINUOUS FUNCTIONS
f ( x ) =
4 , x = 0;
— x + 4 x + 2 , 1 ≤ x < 3 ;
f ( x ) =
Example 4.3. Classify the points of discontinuity of the function
x^2 , — 2 ≤ x < 0;
x
, 0 < x ≤ 2_._
Solution. The function f ( x ) is defined on the interval [ 2 , 2 ]. Since the function
x
2 is continuous on the interval [ 2 , 0 ) and the function 1 /x is continuous on
( 0 , 2 ] , the only point that needs to be considered is x = 0.
lim x → 0 —
f ( x ) = lim x → 0 —
x
2 = 0 ; lim x → 0 +
f ( x ) = lim x → 0 +^ x^
Since the right limit does not exist (equals infinity), f ( x ) has a type II disconti-
nuity at x = 0.
Example 4.4. Classify the points of discontinuity of the function
0 , x < 0;
x, 0 x < 1;
2
4 — x, x ≥ 3_._
Solution. The points that are possible points of discontinuity are x = 0 , x = 1 , and
x = 3.
lim f ( x ) = lim 0 = 0 ; lim f ( x ) = lim x = 0. Since f ( 0 ) = 0 as well, x → 0 —^ x → 0 —^ x → 0 +^ x → 0 +
f ( x ) is continuous at x = 0.
lim x → 1 —^
f ( x ) = lim x → 1 —^
x = 1 ; lim x → 1 +^
f ( x ) = lim x → 1 +^
(— x
2
- 4 x + 2 ) = 5. Since the left
and right limits of f ( x ) exist but are not equal to each other, f ( x ) has a type I
discontinuity at x = 1.
lim x → 3 —^
f ( x ) = lim x → 3 —^
( x^2 + 4 x + 2 ) = 5 ; lim x → 3 +^
f ( x ) = lim x → 3 +^
( 4 — x ) = 1. Since the
left and right limits of f ( x ) exist but are not equal to each other, f ( x ) has a type I
discontinuity at x = 3.
y
In particular, if one of the one-sided limits is in-
finite, then f has a Type II discontinuity at that
point. However, this is only a special case of a
Type II discontinuity.
0 a^
x
CHAPTER 4. CONTINUOUS FUNCTIONS 25
x^2
x
2 — 4 4.35. f ( x (^) ) = ( x
If f is continuous on the closed interval [ a, b ] , then it is bounded on [ a, b ] ,
i.e. there exist m and M such that m ≤ f ( x ) ≤ M for all x ∈ [ a, b ].
If f is continuous on the closed interval [ a, b ] , then f attains its minimum and
x ∈ [ a, b ] , and there exists c 2 ∈ [ a, b ] such that f ( x ) ≤ f ( c 2 ) , x ∈ [ a, b ].
If f is continuous on the closed interval [ a, b ] , m is its minimum value and
M is its maximum value on [ a, b ] , then for any μ , m < μ < M , there exists
c ∈ [ a, b ] such that f ( c ) = μ.
maximum values on [ a, b ] , i.e. there exists c 1 ∈ [ a, b ] such that f ( x ) ≥ f ( c 1 ) ,
A special case of the intermediate value theorem is the
Theorem (Root theorem)
If f is continuous on the closed interval [ a, b ] and its values at the end-
points of the interval have different signs (i.e. f ( a ) f ( b ) < 0 ), then there exists
c ∈ ( a, b ) such that f ( c ) = 0.
Find the points of discontinuity and classify them.
4.31. f ( x ) =
x
| x |
f ( x ) =
x + 4
4.33. f
( x ) = ( x + 4 )^2 (
4.34. f ( x ) =
x
2
| x — 2 |
4.3 Properties of continuous functions
4.52. Prove that if f ( x ) is continuous on ( a, b ) and x 1 , x 2 and x 3 belong to
( a, b ), then there exists c ( a, b ) such that f ( c ) =
1 ( f ( x 1 ) + f ( x 2 ) + f ( x 3 )). 4.53. Prove that any polynomial with an odd highest power has at least one root.
4 + x, x ≤ 1 , 4.36. f
28 CHAPTER 5. DERIVATIVES
Example 5.1. Find the derivative of f ( x ) = 4 x^2 using the definition.
Solution.
f ( x + Δ x )— f ( x ) = 4 ( x + Δ x )^2 — 4 x^2 = 4 x^2 + 8 x Δ x + 4 (Δ x )^2 — 4 x^2 =
= 8 x Δ x + 4 (Δ x )
2 .
Therefore,
f ′^ x lim
f ( x + Δ x )— f ( x ) lim
8 x Δ x + 4 (Δ x )
2
( ) = lim^8 x^^4 x^^8 x Δ x → 0
Δ x Δ x → 0
Δ x Δ x → 0
Find the derivatives of the following functions using the definition.
5.1. x 5.2.
x
5.3. x
3
5. 4. x
2
x 5.6. 4 sin
x
Approximate calculation of the derivative
If it is not possible to find the exact expression for the derivative (for instance, if f ( x ) is defined by a graph or by a table), then the approximate value of f ′( x )
at the point x = x 0 equals
where Δ x = x — x 0. Remember that Δ x does not always have to be positive. In practice, you can use any of the following formulas:
f ( x 0 + h )— f ( x 0 )
h
f ( x 0 )— f ( x 0 — h )
h
f ( x 0 + h )— f ( x 0 — h )
2 h
The second derivative can be approximated by using the formula
f ′′( x 0 ) ≈
f ( x 0 + h )— 2 f ( x 0 ) + f ( x 0 — h )
h^2
30 CHAPTER 5. DERIVATIVES
Example 5.3. Find the derivatives of the following functions:
a) y = x^3 tan—^1 x ; b) y =
x
2
1 + 2 tan x.
Solution. a) Using the product rule,
x
3 tan—
1 x ′^ = ( x
3 )′^ tan—
1 x + x
3 tan—
1 x ′^ = 3 x
2 tan—
1 x + 1
b) Using the quotient rule,
x^3 .
x
2
x^3 + 1
( x
2
3
2
3
( x^3 + 1 )^2
( 2 x + 1 )( x^3 + 1 ) ( x^2 + x 1 ) 3 x^2
( x^3 + 1 )^2
x
4 2 x
3
2
( x^3 + 1 )^2
c) Here it is necessary to use the chain rule. The “outer” function is
x ; the
“inner” function is 1 + 2 tan x. Their derivatives are, respectively, 1 / ( 2
x ) and 2 / cos
2 x. Therefore, bearing in mind that the argument of the outer function is
1 + 2 tan x , ′ 1 2 1 √ 1 + 2 tan x =^ 2
1 + 2 tan x
cos^2 x
cos^2 x
1 + 2 tan x
Example 5.4. Find the derivative of x
x .
Solution. This function cannot be differentiated in this form. It is first necessary to
rewrite it, in order to be able to use the table of elementary derivatives and the rules
- (the quotient rule)
f ( x ) ′^ ′^ g ′( ;
Table of derivatives
- (the chain rule) f ( g ( x ))
′ = f ′ g ( g ( x )) g ′ x ( x ).
- ( x
n )′^ = nx
n — 1
- (log a x )′^ = x ln a
- (cos x )′^ = — sin x
(ln x )′^ = x
( a
x )′^ = a
x ln a , ( e
x )′^ = e
x
- (sin x )′^ = cos x
- (tan x )′^ = cos^2 x 1
- (cot x )′^ =
sin
2 x
- sin—^ x =
- cos—^ x
- tan—
1 x
′
1 ′^ √ x^2 1
1 + x^2
11 cot—^1 x
′
1 + x^2
CHAPTER 5. DERIVATIVES 31
Definition. The inverse function of y = f ( x ) is the function x = f^ −
1 ( y ); in other words,
f^ −
1 ( f ( x )) = x, or f ( f^ −
1 ( x ) = x.
of differentiation:
Therefore,
xx^ = e ln( x
x ) = ex^ ln^ x
( xx )′^ = ex ln x^
′ = ex ln x^ ( x ln x )′^ = ex ln x^ (ln x + 1 ) = xx^ (ln x + 1 ).
Find the derivatives of the following functions.
x
2
2 x^2 + 4 x + 3
( x^2 — 4 )^4
5.23.^3 ( x (^3) + 1 ) 2
x
5.22. ( x
2
x^2 — 3
3 x + 2
5.25. √ 3 x^3 + 1
2 — 3 x
5.40. sin( x
2
sin
2 x
1 + sin^2 x
tan^9 x +
tan^5 x 5.43. x — tan x +
tan^3 x
9 5 3
5.91. Prove that the function y = − ln( x + 1 ) satisfies the equation
xy^ ′^ + 1 = ey.
5.3 Differentiation of inverse functions
CHAPTER 5. DERIVATIVES 35
y
Definition. An implicit function is defined by the equation F ( x, y ) = 0.
The derivative of an implicit function can be found using the equality
dx
[ F ( x, y ( x ))] = 0_._
d
5.4 Implicit differentiation
Example 5.7. Find the derivative of the functions
a) x^2 + y^2 — a^2 = 0 ; b) y^6 — y — x^2 = 0 ; c) y — x —
sin y = 0.
Solution. a) Differentiating by x ,
2 x + 2 yy ′^ = 0 ⇒ y ′^ = —
x .
b) Differentiating by x ,
6 y^5 y ′^ y ′^2 x 0 y ′^
2 x .
6 y^5 — 1
c) Differentiating by x ,
y ′^ — 1 —
(cos y ) y ′^ = 0 ⇒ y ′^ =
4 4 — cos y
Find the derivative of the following implicit functions. 2 2 5.104. x^2 + 2 xy y^2 2 x 5.105.
x y 1 — = a^2
b^2
5.106. x
3
3 — 3 axy = 0 5.107. y
3 — 3 y + 2 ax = 0
5.113. Prove that the function y ( x ) defined by the equation xy ln y = 1 also
satisfies the equation
y^2 + ( xy 1 )
dy = 0_. dx_
36 CHAPTER 5. DERIVATIVES
y
The tangent line:
y = f ′( x 0 )( x — x 0 ) + f ( x 0 )
x
y
The normal line:
y = — f ′( x 0 )
( x — x 0 ) + f ( x 0 )
x
y
If f ′( x 0 ) = 0 , then the graph of f ( x ) has a hori-
zontal tangent line and a vertical normal line at x 0.
y
x
If lim f ′( x ) = ∞ , then the graph of f ( x ) has a ver-
tical tangent line and a horizontal normal line at x 0. Note that in this case f ( x ) is not differentiable
at x = x 0.
x → x 0
0 x 0
x
Definition. The function f ( x ) is smooth at x = x 0 if the graph of f ( x ) has a
unique tangent line.
Note that a differentiable function is always smooth, but a smooth function
can have a vertical tangent line—and therefore it is not differentiable at that point.
5.5 Tangent and normal lines
Example 5.8. Find the equations of the tangent line and of the normal line drawn
to the graph of y = x
3 at the point M ( 1 , 1 ).
Solution. y ′( x ) = 3 x^2 , so y ′( 1 ) = 3.
The tangent line: y = 3 ( x — 1 ) + 1 = 3 x — 2.
The normal line: y = —
( x — 1 ) + 1 = —
x +
38 CHAPTER 5. DERIVATIVES
6 100
Theorem (Rolle’s Theorem)
If f ( x ) is continuous on [ a, b ] , differentiable on ( a, b ), and f ( a ) = f ( b ), then
there exists c ∈ ( a, b ) such that f^ ′( c ) = 0.
y The geometrical interpretation of Rolle’s Theorem
is that if a function is differentiable and assumes
the same value at the ends of an interval, then there
is a point where the tangent line drawn to the graph
of f ( x ) is horizontal.
0 a^ c^
x b
If f ( x ) is continuous on [ a, b ] , differentiable on ( a, b ), then there exists
f ( b ) − f ( a )
c ∈ ( a, b ) such that f^ ′( c ) =
b − a
y
The geometrical interpretation of the Mean Value
Theorem can be given as follows. If a secant line
is drawn between any two points on the graph of
a differentiable function, there exists a point on the
graph between these two points at which the tan-
gent line to the graph of f is parallel to the secant. 0 a^ c^
x b
Using differentials, find the approximate value of the following expressions. Com-
pare your results to the exact values.
5.137. log 11 5.138. sin π^ + π
5.7 Rolle’s Theorem and the Mean Value Theorem
5.141. Check the validity of Rolle’s Theorem for the function
f ( x ) = ( x − 1 )( x − 2 )( x − 3 ).
5.142. Check the validity of Rolle’s Theorem for the function
on the interval [ 1 , 2 ].
f ( x ) =
3
x^2 − 3 x + 2
Find the number in the given interval that satisfies the conclusion of the Mean
Value Theorem.
5.147. f ( x ) = x^2 − 5 x + 7, x ∈ [ − 1 , 3 ] 5.148. f ( x ) = x^3 − 6 x^2 + 9 x +2, x ∈ [ 0 , 4 ]
∞
x →+∞ (^) ex^
x →+∞ (^) ex^
x →+∞ (^) ex^
→
One of the most important methods for calculating limits is L’Hospital’s rule.
I. The indeterminate forms
and
If lim
f ( x ) cannot be found directly, such as when (1) lim f ( x ) = 0 and
x → a
x ) = 0 , giving rise to the indeterminate form , or (2) lim f ( x ) = ∞ and
0
x → a
x → a
lim g ( x ) = ∞, giving rise to the indeterminate form
0 ∞
x → a
∞ ,^ then
lim
f ( x ) = lim
f ′( x ) ,
assuming that the second limit exists or equals infinity. If necessary, L’Hospital’s rule can be used several times in succession. Note also that L’Hospital’s rule
remains valid for x → ∞.
Remember that L’Hospital’s rule can only be used for indeterminate forms!
Chapter 6.
APPLICATIONS OF THE DERIVATIVE
6.1 L’Hospital’s Rule
x
3 — 1 π^ —^ tan—
(^1) x x
2
Example 6.1. Find a) lim ; b) lim
2 1 ;^ c)^ lim^ x. x → (^1) ln x x →+∞ (^) ln( 1 + x^2
) x →+∞^ e
Solution. a) Since lim ( x
3 — 1 ) = 0 and lim ln x = 0 , we can use L’Hospitals’ rule: x → 1
lim
x
3 — 1
x → 1
lim
3 x
2
lim 3 x
3 3
x → 1
ln x x → 1 1 x^^1
x
b) Since lim x →+∞
tan—^1 x =
π , it is again necessary to use L’Hospital’s rule: 2 π (^) — tan— (^1) x — 1
1 2
1 x^2 + 1 x^3 lim
2 1 =^ lim^
x →+∞ (^) ln( 1 + x^2 )^
x →+∞ 1
1 x^2
x →+∞ (^1) + x x 2
c) Here we have an indeterminate form of the type ∞^ , and L’Hospital’s rule
can be used. Note that it is necessary to use L’Hospital’s rule twice:
lim
x^2 lim
2 x lim
CHAPTER 6. APPLICATIONS OF THE DERIVATIVE 43
1
sin x sin x
ex^ — 1
x
Example 6.3. Find lim x → 0
ex —
1 sin x .
Solution. This expression is an indeterminate form of the type ( 1 ∞) at x = 0 , because lim e
x — 1 = 1 and lim 1 = ∞. x → 0 x^ x → 0 sin^ x Transforming the expression gives
e
x — 1
1 ln( ex x — (^1) ) (^) lim ln( ex — 1 )—ln x lim x → 0 x
= lim e x → 0
sin x (^) = ex →^0_._
Thus, the problem reduces to finding the limit
lim ln( e
x (^) — 1 )— ln x = lim
ex ex — 1
1 x = lim
xex^ ex^ + 1
x → 0 sin x x → 0 cos x x → 0 x ( ex^ — 1 ) e
x
x — e
x xe
x 1 1 1 = lim (^) x x = lim (^) x x = lim (^) ex — 1 = =.
x → 0 e (^) + xe — 1 x → 0 xe (^) + e — (^1) x → (^0 1) + (^) xex 1 + 1 2
Therefore, the answer is
lim x → 0
1 sin x =
e. x
Find the following limits using L’Hospital’s rule.
6.1. lim
x
3 — 3 x
2
x → 1 x^3 — 4 x^2 + 3 tan x — x
6.2. lim
sin ax
x → 0 sin 3 b x tan x — 1 6.3. lim x → 0 x — sin x
6.4. lim x → π/ 4 2 sin^2 x — 1
6.5. lim
ax^ — xa
6.6. lim
ln(sin ax )
x → a x — a x → 0 ln(sin bx )
III. The indeterminate form ( 1
∞ ).
x → a x → a ∞
x → a
turn can be reduced to the form (^0 ) or (∞^ )) by using the properties of the
This problem can be reduced to the indeterminate form ( 0 · ∞) (which in its
exponential function:
0 ∞
f ( x ) g ( x )^ = e ln^ f^ ( x )
g ( x ) = eg ( x )^ ln^ f^ ( x ).
(The function g ( x ) approaches infinity, while f ( x ) approaches 1 and so ln f ( x )
approaches 0 .)
44 CHAPTER 6. APPLICATIONS OF THE DERIVATIVE
5 4
Definition. The function f ( x ) is strictly increasing on ( a, b ), if for any points
x 1 and x 2 ( x 1 < x 2 ) on this interval we have f ( x 1 ) < f ( x 2 ).
Definition. The function f ( x ) is strictly decreasing on ( a, b ), if for any points
x 1 and x 2 ( x 1 < x 2 ) on this interval we have f ( x 1 ) > f ( x 2 ).
Theorem. If the differentiable function f ( x ) is strictly increasing on ( a, b ),
then f ′( x ) ≥ 0 for all x ∈ ( a, b ).
Theorem. If f ′( x ) > 0 for all x ∈ ( a, b ), then f ( x ) is strictly increasing on ( a, b ).
Analogous theorems can be proven for decreasing functions.
Important: Note that a strictly increasing function can have a zero derivative at
isolated points. This behavior is exhibited, for instance, by the function f ( x ) = x^3.
This function is strictly increasing for all x , and yet f^ ′( 0 ) = 0.
The same is true for strictly decreasing functions.
6.2 Monotonicity
Example 6.4. Find the intervals on which the function f ( x ) = 3 x
4 4 x
3 12 x
2
- 2 is increasing and decreasing.
Solution. First find the derivative:
f^ ′( x ) = 12 x^3 − 12 x^2 − 24 x = 12 x ( x^2 − x − 1 ) = 12 x ( x + 1 )( x − 2 ).
Critical points: f^ ′( x ) = 0 or f^ ′( x ) /∃ :
12 x ( x + 1 )( x − 2 ) = 0 ⇒ x 1 = 0; x 2 = − 1 , x 3 = 2.
Check the sign of the derivative:
x
We see that f^ ′( x ) is positive for x ∈ ( − 1 , 0 ) and for x ∈ ( 2 , ∞), while f^ ′( x )
is negative for x ∈ ( −∞ , − 1 ) and x ∈ ( 0 , 2 ). Therefore, f ( x ) is increasing for
x ∈ ( − 1 , 0 ) ∪ ( 2 , +∞) and decreasing for x ∈ ( −∞ , − 1 ) ∪ ( 0 , 2 ).
Find the intervals on which the functions are strictly increasing or decreasing.
6.32. f ( x ) = 3 x − x^3 6.33. f ( x ) = x^4 − 2 x^2 − 5
2 x
6.34. f ( x ) = ( x − 2 ) ( 2 x + 1 ) 6.35. f ( x ) =
6.36. f ( x ) =
x
x + 100
1 + x^2
6.37. f ( x ) = x − e
x
f^ ·( x ) — 0
|
|
|
f ( x ) − 1 0 2
