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Chapter 4.
CONTINUOUS FUNCTIONS
4.1 Continuity
Definition. The function f is continuous at x = a if lim x→a f (x) = f (a).
This definition is equivalent to the condition lim x→a−^
f (x) = lim x→a+^
f (x) = f (a).
Intuitively, continuous functions have a graph which can be drawn without lifting the pencil.
Example 4.1. Define f ( 1 ) so that the function f (x) =
x^2 − 1 x^3 − 1
is continuous at x = 1. Solution. By the definition, f (x) will be continuous at x = 1 if lim x→ 1
f (x) = f ( 1 ).
lim x→ 1
f (x) = lim x→ 1
x^2 − 1 x^3 − 1
= lim x→ 1
x + 1 x^2 + x + 1
Therefore, f (x) will be continuous at x = 1 if f ( 1 ) = 23.
Define f ( 0 ) so that the following functions will be continuous at x = 0.
4.1. f (x) =
1 + x − 1 x
. 4.2. f (x) =
1 + 2 x − 1 √ (^31) + 2 x − 1.
4.3. f (x) =
sin x x
. 4.4. f (x) = 4
1 − cos x x^2
4.5. f (x) =
tan 2x x
. 4.6. f (x) = sin x sin
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 6 = 4 , what is f ( 4 )?
4.10. If possible, define f ( 1 ) so that the function f (x) =
x^2 − 2 x + 1 x^2 − 4 x + 3
is contin-
uous at x = 1.
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
is continuous. Solution. Consider the point x = 2. lim x→ 2 −^
f (x) = lim x→ 2 −^
( 3 kx − 5 ) = 6 k − 5;
lim x→ 2 +^
f (x) = lim x→ 2 +^
( 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 = 1311.
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 6 = 3 ,
A, x = 3.
4.14. f (x) =
e^2 x, x < 0 , x − a, x ≥ 0.
4.15. f (x) =
x^2 + 3 x, x ≤ 2 , bx ln x, x > 2.
4.16. f (x) =
e^2 x+d^ , x ≥ 0 , x + 2 , x < 0.
24 CHAPTER 4. CONTINUOUS FUNCTIONS
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.
x
y
a
Example 4.3. Classify the points of discontinuity of the function
f (x) =
x^2 , − 2 ≤ x < 0; 4 , x = 0; 1 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
f (x) =
0 , x < 0; x, 0 ≤ x < 1; − x^2 + 4 x + 2 , 1 ≤ x < 3; 4 − x, x ≥ 3.
Solution. The points that are possible points of discontinuity are x = 0 , x = 1 , and x = 3. lim x→ 0 −^
f (x) = lim x→ 0 −^
0 = 0 ; lim x→ 0 +^
f (x) = lim x→ 0 +^
x = 0. Since f ( 0 ) = 0 as well, 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.
CHAPTER 4. CONTINUOUS FUNCTIONS 25
Find the points of discontinuity and classify them.
4.31. f (x) =
x |x|
4.32. f (x) =
x + 4
4.33. f (x) =
(x + 4 )^2
4.34. f (x) =
x^2 + x − 6 |x − 2 |
4.35. f (x) =
4 + x, x ≤ 1 , 4.36. f (x) =
x^2 − 4 x^2
4.3 Properties of continuous functions
Theorem (Boundedness theorem) 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]. Theorem (Extreme Value theorem) If f is continuous on the closed interval [a, b] , then f attains its minimum and maximum values on [a, b] , i.e. there exists c 1 ∈ [a, b] such that f (x) ≥ f (c 1 ) , x ∈ [a, b] , and there exists c 2 ∈ [a, b] such that f (x) ≤ f (c 2 ) , x ∈ [a, b]. Theorem (Intermediate Value theorem) 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.
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.
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) = 13 ( 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.
28 CHAPTER 5. DERIVATIVES
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
f ′(x 0 ) ≈
△ f △x
f (x 0 + △x) − f (x 0 ) △x
where △x = x − x 0. Remember that △x does not always have to be positive. In practice, you can use any of the following formulas:
f ′(x 0 ) ≈
f (x 0 + h) − f (x 0 ) h
f ′(x 0 ) ≈
f (x 0 ) − f (x 0 − h) h
f ′(x 0 ) ≈
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
Example 5.1. Find the derivative of f (x) = 4 x^2 using the definition. Solution. f (x + △x) − f (x) = 4 (x + △x)^2 − 4 x^2 = 4 x^2 + 8 x△x + 4 (△x)^2 − 4 x^2 = = 8 x△x + 4 (△x)^2. Therefore,
f ′(x) = lim △x→ 0
f (x + △x) − f (x) △x
= lim △x→ 0
8 x△x + 4 (△x)^2 △x
= lim △x→ 0
( 8 x + 4 △x) = 8 x.
Find the derivatives of the following functions using the definition.
5.1. x 5.2.
x 5.3. x^3 5.4. x^2 + 2 x + 2
x 5.6. 4 sin
x 2
CHAPTER 5. DERIVATIVES 29
Example 5.2. Find an approximate value of f ′( 2 ) and f ′′( 2 ) if it is known that f ( 1. 8 ) = 3. 3 , f ( 2 ) = 3. 5 and f ( 2. 2 ) = 3. 6. Solution. There are three equally valid approximate values of f ′( 2 ) that can be found:
f ′( 2 ) ≈
f ( 2. 2 ) − f ( 2 )
- 2 − 2
f ′( 2 ) ≈
f ( 2 ) − f ( 1. 8 ) 2 − 1. 8
f ′( 2 ) ≈
f ( 2. 2 ) − f ( 1. 8 )
- 2 − 1. 8
Remember that we have no reason to consider any of these approximate values to be more accurate than the rest. Now find an approximation for f ′′( 2 ) :
f ′′( 2 ) ≈
f ( 2. 2 ) − 2 f ( 2 ) + f ( 1. 8 )
. 22
5.11. Use the table to find an approximate value of f ′( 3 ) and f ′′( 3 ).
x 2. 95 3 3. 05
f (x) 1. 07 1. 12 1. 18
x
y
5.12. The graph of f (x) , which consists of three line segments, is shown above. Find f ′(− 1. 5 ) , f ′( 1 ) , and f ′( 2. 5 ).
5.2 Differentiation of explicit functions
Rules of differentiation
k f (x)
= k f ′(x) , where k is a constant;
- (the sum rule)
f (x) + g(x)
= f ′(x) + g′(x) ;
CHAPTER 5. DERIVATIVES 31
of differentiation: xx^ = eln(x x) = ex^ ln^ x Therefore, (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 − 4 )^4
x^2 + 3 x + 2 2 x^2 + 4 x + 3
5.22. (x^2 + 6 )
5.23.^3 x^2 −^3
(x^3 + 1 )^2
5.24.
3 x + 2 √ 5.25. (^) 2 − 3 x
x √ (^3) x (^3) + 1
5.40. sin(x^2
tan^9 x +
tan^5
tan^3 x
sin^2 x 1 + sin^2 x
5.91. Prove that the function y = − ln(x + 1 ) satisfies the equation
xy ′^ + 1 = ey.
x 5.43. x − tan x +
5.3 Differentiation of inverse functions
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.
34 CHAPTER 5. DERIVATIVES
Theorem. Suppose that f is a function which is differentiable on an open interval containing x 0. If either f ′(x) > 0 or f ′(x) < 0 for all x in this interval, then f has a differentiable inverse f −^1 at y 0 = f (x 0 ) , and
d f −^1 dy
y=y 0
d f dx
x=x 0
, or x′(y 0 ) =
y′(x 0 )
The condition that f ′(x) > 0 or f ′(x) < 0 for all x in an interval is actually a way of making sure that f (x) has two required properties: it is differentiable and one-to-one, i.e., for every value of y there can only be one value of x.
Example 5.5. Find the derivative of a) y = loga x and b) y = sin−^1 x using the formula for the derivative of the inverse function. Solution. a) The inverse function for y = loga x is x = ay^ , and its derivative is x′(y) = ay^ ln a. Therefore, we have
y′(x) =
x′(y)
ay^ ln a
x ln a
b) The inverse function for y = sin−^1 x is x = sin y , and its derivative is x′(y) = cos y. We should note, however, that it is necessary to restrict the domain of the function x = sin y so that we have a one-to-one function; the standard choice is y ∈
[
−π 2 , π 2
]
According to the theorem,
( sin−^1 x
(sin y)′^
cos y
1 − sin^2 y
1 − x^2
Note that due to the choice of y we have cos y ≥ 0 , and it was for this reason that we put cos y =
1 − sin^2 y.
Example 5.6. Find the derivative of f −^1 (x) at x = 22 , if f (x) = x^4 + x^3 − x and f −^1 ( 22 ) = 2. Solution. Denote x = 2 , y = 22 (so that 22 = 24 + 23 − 2 ). According to the for- mula, d f −^1 dy
y= 22
d f dx
x= 2
4 x^3 + 3 x^2 − 1
x= 2
Find the derivative of f −^1 (x) at x = 3.
5.100. f (x) = 2 x^3 + 1 5.101. f (x) = x + cos x + 2
5.102. f (x) = x + sin x + 3 5.103. f (x) = 4 − log 2 x
36 CHAPTER 5. DERIVATIVES
5.5 Tangent and normal lines
The tangent line: y = f ′(x 0 )(x − x 0 ) + f (x 0 )
x
y
x 0
The normal line:
y = −
f ′(x 0 )
(x − x 0 ) + f (x 0 )
x
y
x 0
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.
x
y
x 0
If lim x→x 0 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
y
0 x 0 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.
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 +
CHAPTER 5. DERIVATIVES 37
Example 5.9. Find the equation of the tangent line drawn to the graph of y = x^2 that is a) parallel to the line y = 4 x − 5 ; b) perpendicular to the line 2 x − 6 y + 5 = 0. Solution. a) Parallel lines have equal slopes; the slope of the tangent line at x 0 is given by y ′(x 0 ) = 2 x 0. Therefore, 2 x 0 = 4 ⇒ x 0 = 2. The equation of the tangent line at x 0 = 2 is y = 4 (x − 2 ) + 4 , or y = 4 x − 4. b) The product of the slopes of perpendicular lines is − 1. Therefore,
2 x 0 ·
= − 1 ⇒ x 0 = −
The equation of the tangent line is y = − 3
x +
, or y = − 3 x −
5.115. Find the points on the curve y = x^2 (x− 2 )^2 where the tangent line is parallel to the x -axis.
5.116. Find the equations of the tangent lines drawn to the graph of y = x − (^1) x at the x -intercepts.
5.117. Find the equation of the tangent line drawn to the graph of y = sin x at the
point
3 π 4 ,
√ 2 2
5.118. Find the equation of the tangent line drawn to the graph of y = sin x at the point (x 0 , y 0 ). 3
5.6 The differential
Definition. The differential of f (x) is given by d f = f ′(x)dx. If x is sufficiently close to x 0 and f (x 0 ) is known, then differentials can be used to find an approximate value of f (x) :
f (x) ≈ f (x 0 ) + d f = f (x 0 ) + f ′(x 0 )(x − x 0 ).
Note that this can also be understood as the tangent line approximation of f (x) at x = x 0.
Example 5.10. Using differentials, find the approximate value of 20. 12.
Solution. Consider the function f (x) = x^2. The differential of this function is d f = 2 xdx. At x 0 = 20 , f (x 0 ) = 400 and f ′(x 0 ) = 40. Therefore,
- 12 ≈ 400 + 40 · 0. 1 = 404.
Chapter 6.
APPLICATIONS OF THE DERIVATIVE
6.1 L’Hospital’s Rule
One of the most important methods for calculating limits is L’Hospital’s rule.
I. The indeterminate forms
and
If lim x→a
f (x) g(x)
cannot be found directly, such as when (1) lim x→a f (x) = 0 and
lim x→a g(x) = 0 , giving rise to the indeterminate form
0
, or (2) lim x→a f (x) = ∞ and lim x→a g(x) = ∞ , giving rise to the indeterminate form
∞
, then
lim x→a
f (x) g(x)
= lim x→a
f ′(x) g′(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!
Example 6.1. Find a) lim x→ 1
x^3 − 1 ln x
; b) lim x→+∞
π 2 −^ tan
− (^1) x
ln( 1 + (^) x^12 )
; c) lim x→+∞
x^2 ex^
Solution. a) Since lim x→ 1
(x^3 − 1 ) = 0 and lim x→ 1
ln x = 0 , we can use L’Hospitals’ rule:
lim x→ 1
x^3 − 1 ln x
= lim x→ 1
3 x^2 1 x
= lim x→ 1
3 x^3 = 3.
b) Since lim x→+∞
tan−^1 x =
π 2
, it is again necessary to use L’Hospital’s rule:
lim x→+∞
π 2 −^ tan
− (^1) x
ln( 1 + (^) x^12 )
= lim x→+∞
− (^1) +^1 x 2 1 1 + 1 /x^2
− (^) x^23
) (^) = lim x→+∞
1 + x^2
x^2 + 1 x^2
x^3 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→+∞
x^2 ex^
= lim x→+∞
2 x ex^
= lim x→+∞
ex^
[
]
42 CHAPTER 6. APPLICATIONS OF THE DERIVATIVE
II. The indeterminate forms ( 0 · ∞) and (∞ − ∞).
If lim x→a
f (x) = 0 and lim x→a
g(x) = ∞ , then finding lim x→a
f (x)g(x) involves dealing with the indeterminate form ( 0 · ∞). If lim x→a f (x) = +∞ and lim x→a g(x) = +∞ , then the limit lim x→a
f (x) − g(x)
is also indeterminate and can be expressed as (∞ − ∞). Note that this limit will also be indeterminate if lim x→a f (x) = −∞ and lim x→a g(x) = −∞. On the other hand, if lim x→a f (x) = +∞ and lim x→a g(x) = −∞ , then lim x→a
f (x) − g(x)
= (+∞ − (−∞)) = +∞ , and this limit is not an indeterminate form. These limits can be reduced to the indeterminate forms
0
or
∞
by alge- braic transformations, after which they can be calculated using L’Hospital’s rule.
Example 6.2. Find the following limits:
a) lim x→π/ 2
x −
π 2
tan x ; b) lim x→ 1
ln x
x − 1
; c) lim x→+∞
ex^ − x^2
Solution. a) It is enough to use trigonometric transformations here:
lim x→π/ 2
x −
π 2
tan x = lim x→π/ 2
x −
π 2 cot x
= lim x→π/ 2
− (^) sin^12 x
= − lim x→π/ 2
sin^2 x = − 1.
b) Simplifying,
lim x→ 1
ln x
x − 1
= lim x→ 1
x − 1 − ln x (x − 1 ) ln x
= lim x→ 1
1 − (^1) x ln x + x− x^1
= lim x→ 1
x − 1 x ln x + x − 1
Using L’Hospital’s rule a second time,
lim x→ 1
x − 1 x ln x + x − 1
= lim x→ 1
ln x + 1 + 1
Therefore, lim x→ 1
ln x
x − 1
c) We will use algebraic transformations to find this limit:
lim x→+∞
ex^ − x^2
= lim x→+∞
ex
x^2 ex
The limit of the expression in parenthesis is
lim x→+∞
x^2 ex
= 1 − lim x→+∞
2 x ex^
= 1 − lim x→+∞
ex^
Therefore, since lim x→+∞ ex^ = +∞ , lim x→+∞ ex
x^2 ex
= [+∞ · 1 ] = +∞.
44 CHAPTER 6. APPLICATIONS OF THE DERIVATIVE
6.2 Monotonicity
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.
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) 6 ∃ : 12 x(x + 1 )(x − 2 ) = 0 ⇒ x 1 = 0; x 2 = − 1 , x 3 = 2. Check the sign of the derivative:
| | | − 1 0 2
x
f ′( x ) f ( 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
6.34. f (x) = (x − 2 )^5 ( 2 x + 1 )^4 6.35. f (x) =
2 x 1 + x^2
6.36. f (x) =
x x + 100
6.37. f (x) = x − ex
46 CHAPTER 6. APPLICATIONS OF THE DERIVATIVE
6.3 Related rates
Problems in related rates deal with the change of various quantities (physical or geometrical) with time. Definition. The rate of change of a quantity is the derivative of that quantity with respect to time. Always remember to include units of measurement!!
Position, Velocity, Acceleration (P.V.A.)
Definition. The position of a particle moving along the x -axis is given by the function x(t).
Definition. The velocity of a particle moving along the x -axis is given by
dx dt
Definition. The speed of a particle moving along the x -axis is given by
dx dt
Definition. The acceleration of a particle moving along the x -axis is given by d^2 x dt^2
or
dv dt
Analogous definitions are valid for movement along the y -axis.
Example 6.5. A rectangle has sides of 20 and 40 inches, respectively. The larger sides of the rectangle begin to shrink at a rate of 2 inches per second. How fast is the area of the rectangle changing at this moment? Solution. Let a be the length of the smaller sides of the rectangle and b be the length of the larger sides. Since b is decreasing , we have db dt
= − 2. (inches per second)
Note that a remains constant, so da dt
= 0. (inches per second)
The area of the rectangle is given by A = ab , so its rate of change is given by dA dt
d dt
(ab) =
da dt
b + a
db dt
= − 2 a. ( in^2 per second)
At the instant when a = 20 inches and b = 40 inches,
dA dt
∣ a=^20 b= 40
= − 40. ( in^2 per second)
Example 6.6. The hypotenuse of a right triangle is increasing at the rate of 4 inches per minute, while all the angles in the triangle remain constant. At the instant when the sides of the triangle are 10 , 10
3 and 20 inches, determine a) how fast the perimeter of the triangle is changing; b) how fast the area of the triangle is changing.
