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(^104) MATHEMATICS
v The whole of science is nothing more than a refinement
of everyday thinking. ” — ALBERT EINSTEIN v
5.1 Introduction
This chapter is essentially a continuation of our study of differentiation of functions in Class XI. We had learnt to differentiate certain functions like polynomial functions and trigonometric functions. In this chapter, we introduce the very important concepts of continuity, differentiability and relations between them. We will also learn differentiation of inverse trigonometric functions. Further, we introduce a new class of functions called exponential and logarithmic functions. These functions lead to powerful techniques of differentiation. We illustrate certain geometrically obvious conditions through differential calculus. In the process, we will learn some fundamental theorems in this area.
5.2 Continuity
We start the section with two informal examples to get a feel of continuity. Consider the function
1, if 0 ( ) 2, if 0
x f x x
^ ≤
This function is of course defined at every point of the real line. Graph of this function is given in the Fig 5.1. One can deduce from the graph that the value of the function at nearby points on x -axis remain close to each other except at x = 0. At the points near and to the left of 0, i.e., at points like – 0.1, – 0.01, – 0.001, the value of the function is 1. At the points near and to the right of 0, i.e., at points like 0.1, 0.01,
Chapter 5
CONTINUITY AND
DIFFERENTIABILITY
Sir Issac Newton (1642-1727)
Fig 5.
CONTINUITY AND DIFFERENTIABILITY 105
0.001, the value of the function is 2. Using the language of left and right hand limits, we may say that the left (respectively right) hand limit of f at 0 is 1 (respectively 2). In particular the left and right hand limits do not coincide. We also observe that the value of the function at x = 0 concides with the left hand limit. Note that when we try to draw the graph, we cannot draw it in one stroke, i.e., without lifting pen from the plane of the paper, we can not draw the graph of this function. In fact, we need to lift the pen when we come to 0 from left. This is one instance of function being not continuous at x = 0. Now, consider the function defined as
f x
x x
if if
This function is also defined at every point. Left and the right hand limits at x = 0 are both equal to 1. But the value of the function at x = 0 equals 2 which does not coincide with the common value of the left and right hand limits. Again, we note that we cannot draw the graph of the function without lifting the pen. This is yet another instance of a function being not continuous at x = 0.
Naively, we may say that a function is continuous at a fixed point if we can draw the graph of the function around that point without lifting the pen from the plane of the paper.
Mathematically, it may be phrased precisely as follows:
Definition 1 Suppose f is a real function on a subset of the real numbers and let c be a point in the domain of f. Then f is continuous at c if
lim ( ) ( ) x c
f x f c →
More elaborately, if the left hand limit, right hand limit and the value of the function at x = c exist and equal to each other, then f is said to be continuous at x = c. Recall that if the right hand and left hand limits at x = c coincide, then we say that the common value is the limit of the function at x = c. Hence we may also rephrase the definition of continuity as follows: a function is continuous at x = c if the function is defined at x = c and if the value of the function at x = c equals the limit of the function at x = c. If f is not continuous at c , we say f is discontinuous at c and c is called a point of discontinuity of f.
Fig 5.
CONTINUITY AND DIFFERENTIABILITY 107
Solution The function is defined at x = 0 and its value at x = 0 is 1. When x ≠ 0, the function is given by a polynomial. Hence,
0
lim ( ) x
f x → =^
3 3 0
lim ( 3) 0 3 3 x
x →
Since the limit of f at x = 0 does not coincide with f (0), the function is not continuous at x = 0. It may be noted that x = 0 is the only point of discontinuity for this function.
Example 5 Check the points where the constant function f ( x ) = k is continuous.
Solution The function is defined at all real numbers and by definition, its value at any real number equals k. Let c be any real number. Then
lim ( ) x c
f x → =^
lim x c
k k →
Since f ( c ) = k = lim x → c f ( x ) for any real number c , the function f is continuous at
every real number.
Example 6 Prove that the identity function on real numbers given by f ( x ) = x is continuous at every real number.
Solution The function is clearly defined at every point and f ( c ) = c for every real number c. Also,
lim ( ) x c
f x →
= lim x → c^ x^^ = c
Thus, lim x → cf ( x ) = c = f ( c ) and hence the function is continuous at every real number.
Having defined continuity of a function at a given point, now we make a natural extension of this definition to discuss continuity of a function.
Definition 2 A real function f is said to be continuous if it is continuous at every point in the domain of f.
This definition requires a bit of elaboration. Suppose f is a function defined on a closed interval [ a , b ], then for f to be continuous, it needs to be continuous at every point in [ a , b ] including the end points a and b. Continuity of f at a means
lim ( ) x a +^ f^ x →
= f ( a )
and continuity of f at b means
x^ lim → b –^^ f^ ( ) x =^ f ( b )
Observe that lim ( ) x a −^ f^ x →
and lim ( ) x b +^ f^ x →
do not make sense. As a consequence
of this definition, if f is defined only at one point, it is continuous there, i.e., if the domain of f is a singleton, f is a continuous function.
(^108) MATHEMATICS
Example 7 Is the function defined by f ( x ) = | x |, a continuous function?
Solution We may rewrite f as
f ( x ) =
, if 0 , if 0
x x x x
^ −^ <
By Example 3, we know that f is continuous at x = 0.
Let c be a real number such that c < 0. Then f ( c ) = – c. Also
lim ( ) x c
f x → =^
lim ( ) – x c
x c →
− = (^) (Why?)
Since lim ( ) ( ) x c
f x f c →
= , f is continuous at all negative real numbers.
Now, let c be a real number such that c > 0. Then f ( c ) = c. Also
lim ( ) x c
f x → =^
lim x c
x c →
= (^) (Why?)
Since lim ( ) ( ) x c
f x f c →
= , f is continuous at all positive real numbers. Hence, f
is continuous at all points.
Example 8 Discuss the continuity of the function f given by f ( x ) = x^3 + x^2 – 1.
Solution Clearly f is defined at every real number c and its value at c is c^3 + c^2 – 1. We also know that
lim ( ) x c
f x → =^
lim ( 3 2 1) 3 2 1 x c
x x c c →
Thus lim ( ) ( ) x c
f x f c →
= , and hence f is continuous at every real number. This means
f is a continuous function.
Example 9 Discuss the continuity of the function f defined by f ( x ) =
x
, x ≠ 0.
Solution Fix any non zero real number c , we have
1 1 lim ( ) lim x c x c
f x → → x c
Also, since for c ≠ 0,
f c ( ) c
= (^) , we have lim ( ) ( ) x c
f x f c →
= and hence, f is continuous
at every point in the domain of f. Thus f is a continuous function.
(^110) MATHEMATICS
Example 10 Discuss the continuity of the function f defined by
f ( x ) =
2, if 1 2, if 1
x x x x
^ +^ ≤
−^ >
Solution The function f is defined at all points of the real line.
Case 1 If c < 1, then f ( c ) = c + 2. Therefore, (^) lim ( ) lim( 2) 2 x c x c
f x x c → →
Thus, f is continuous at all real numbers less than 1.
Case 2 If c > 1, then f ( c ) = c – 2. Therefore,
lim ( ) lim x c x c
f x → →
= ( x – 2) = c – 2 = f ( c )
Thus, f is continuous at all points x > 1.
Case 3 If c = 1, then the left hand limit of f at x = 1 is
x^ lim → 1 –^ f^ ( ) x^^ =^ x lim (→ 1 –^ x +^ 2)^ =^1 +^2 =^3
The right hand limit of f at x = 1 is
1 1
lim ( ) lim ( 2) 1 2 1 x x +^ f^ x^ + x → →
Since the left and right hand limits of f at x = 1 do not coincide, f is not continuous at x = 1. Hence x = 1 is the only point of discontinuity of f. The graph of the function is given in Fig 5.4.
Example 11 Find all the points of discontinuity of the function f defined by
f ( x ) =
2, if 1 0, if 1 2, if 1
x x x x x
^ +^ <
Solution As in the previous example we find that f is continuous at all real numbers x ≠ 1. The left hand limit of f at x = 1 is
x^ lim → 1 −^ f^ ( ) x^^ =^ lim ( x → 1 –^ x +^ 2)^ =^1 +^2 =^3 The right hand limit of f at x = 1 is
1 1
lim ( ) lim ( 2) 1 2 1 x x +^ f^ x^ + x → →
Since, the left and right hand limits of f at x = 1 do not coincide, f is not continuous at x = 1. Hence x = 1 is the only point of discontinuity of f. The graph of the function is given in the Fig 5.5.
Fig 5.
Fig 5.
CONTINUITY AND DIFFERENTIABILITY 111
Example 12 Discuss the continuity of the function defined by
f ( x ) =
2, if 0 2, if 0
x x x x
−^ +^ >
Solution Observe that the function is defined at all real numbers except at 0. Domain of definition of this function is
D 1 ∪ D 2 where D 1 = { x ∈ R : x < 0} and D 2 = { x ∈ R : x > 0}
Case 1 If c ∈ D 1 , then (^) lim ( ) lim x c x c
f x → →
= ( x^ + 2)
= c + 2 = f ( c ) and hence f is continuous in D 1.
Case 2 If c ∈ D 2 , then lim ( ) lim x c x c
f x → →
= ( – x + 2)
= – c + 2 = f ( c ) and hence f is continuous in D 2. Since f is continuous at all points in the domain of f , we deduce that f is continuous. Graph of this function is given in the Fig 5.6. Note that to graph this function we need to lift the pen from the plane of the paper, but we need to do that only for those points where the function is not defined.
Example 13 Discuss the continuity of the function f given by
f ( x ) = (^2)
, if 0 , if 0
x x x x
^ ≥
Solution Clearly the function is defined at every real number. Graph of the function is given in Fig 5.7. By inspection, it seems prudent
to partition the domain of definition of f into three disjoint subsets of the real line.
Let D 1 = { x ∈ R : x < 0}, D 2 = {0} and D 3 = { x ∈ R : x > 0}
Case 1 At any point in D 1 , we have f ( x ) = x^2 and it is easy to see that it is continuous there (see Example 2).
Case 2 At any point in D 3 , we have f ( x ) = x and it is easy to see that it is continuous there (see Example 6).
Fig 5.
Fig 5.
CONTINUITY AND DIFFERENTIABILITY 113
Case 1 Let c be a real number which is not equal to any integer. It is evident from the graph that for all real numbers close to c the value of the function is equal to [ c ]; i.e.,
lim ( ) lim [ ] [ ] x c x c
f x x c → →
= =. Also f ( c ) = [ c ] and hence the function is continuous at all real
numbers not equal to integers.
Case 2 Let c be an integer. Then we can find a sufficiently small real number r > 0 such that [ c – r ] = c – 1 whereas [ c + r ] = c.
This, in terms of limits mean that
lim x → c −
f ( x ) = c – 1, lim x → c +
f ( x ) = c
Since these limits cannot be equal to each other for any c , the function is discontinuous at every integral point.
5.2.1 Algebra of continuous functions
In the previous class, after having understood the concept of limits, we learnt some algebra of limits. Analogously, now we will study some algebra of continuous functions. Since continuity of a function at a point is entirely dictated by the limit of the function at that point, it is reasonable to expect results analogous to the case of limits.
Theorem 1 Suppose f and g be two real functions continuous at a real number c. Then
(1) f + g is continuous at x = c. (2) f – g is continuous at x = c. (3) f. g is continuous at x = c.
(4) f g
is continuous at x = c , (provided g ( c ) ≠ 0).
Proof We are investigating continuity of ( f + g ) at x = c. Clearly it is defined at x = c. We have
lim( ) ( ) x c
f g x →
f x g x →
- (^) (by definition of f + g )
= lim x → c f^ ( ) x^^ +^ lim x → c^ g x ( ) (by the theorem on limits)
= f ( c ) + g ( c ) (as f and g are continuous) = ( f + g ) ( c ) (by definition of f + g )
Hence, f + g is continuous at x = c.
Proofs for the remaining parts are similar and left as an exercise to the reader.
(^114) MATHEMATICS
Remarks
(i) As a special case of (3) above, if f is a constant function, i.e., f ( x ) = λ for some real number λ, then the function (λ. g ) defined by (λ. g ) ( x ) = λ. g ( x ) is also continuous. In particular if λ = – 1, the continuity of f implies continuity of – f. (ii) As a special case of (4) above, if f is the constant function f ( x ) = λ, then the
function (^) g
λ defined by ( ) ( )
x g g x
λ λ = is also continuous wherever g ( x ) ≠ 0. In
particular, the continuity of g implies continuity of
g. The above theorem can be exploited to generate many continuous functions. They also aid in deciding if certain functions are continuous or not. The following examples illustrate this:
Example 16 Prove that every rational function is continuous.
Solution Recall that every rational function f is given by
( ) ( ) , ( ) 0 ( )
p x f x q x q x
where p and q are polynomial functions. The domain of f is all real numbers except points at which q is zero. Since polynomial functions are continuous (Example 14), f is continuous by (4) of Theorem 1.
Example 17 Discuss the continuity of sine function.
Solution To see this we use the following facts
0
lim sin 0 x
x →
We have not proved it, but is intuitively clear from the graph of sin x near 0. Now, observe that f ( x ) = sin x is defined for every real number. Let c be a real number. Put x = c + h. If x → c we know that h → 0. Therefore
lim ( ) x c
f x → =^
lim sin x c
x →
= lim sin( h → 0 c^ + h )
= lim [sin h → 0 c^ cos^ h^ +cos^ c^ sin^ h ]
= lim [sin h → 0^ c^ cos^ h ]^^ +lim [cos h → 0 c^ sin^ h ] = sin c + 0 = sin c = f ( c ) Thus (^) lim x → c
f ( x ) = f ( c ) and hence f is a continuous function.
(^116) MATHEMATICS
EXERCISE 5.
1. Prove that the function f ( x ) = 5 x – 3 is continuous at x = 0, at x = – 3 and at x = 5. 2. Examine the continuity of the function f ( x ) = 2 x^2 – 1 at x = 3. 3. Examine the following functions for continuity.
(a) f ( x ) = x – 5 (b) f ( x ) =
x − 5
, x ≠ 5
(c) f ( x ) =
x x
, x ≠ –5 (d) f ( x ) = | x – 5 |
4. Prove that the function f ( x ) = xn^ is continuous at x = n , where n is a positive integer. 5. Is the function f defined by , if 1 ( ) 5, if > 1
x x f x x
^ ≤
continuous at x = 0? At x = 1? At x = 2?
Find all points of discontinuity of f , where f is defined by
2 3, if 2 ( ) 2 3, if > 2
x x f x x x
^ +^ ≤
| | 3, if 3 ( ) 2 , if 3 < 3 6 2, if 3
x x f x x x x x
^ +^ ≤ −
, if 0 ( ) 0, if 0
x x f x (^) x x
, if 0 ( )^ |^ | 1, if 0
x x f x x x
1, if 1 ( ) 1, if 1
x x f x x x
^ +^ ≥
+^ < 11.
3 2
3, if 2 ( ) 1, if 2
x x f x x x
+^ >
10 2
1, if 1 ( ) , if 1
x x f x x x
13. Is the function defined by 5, if 1 ( ) 5, if 1
x x f x x x
^ +^ ≤
−^ >
a continuous function?
CONTINUITY AND DIFFERENTIABILITY 117
Discuss the continuity of the function f , where f is defined by
3, if 0 1 ( ) 4, if 1 3 5, if 3 10
x f x x x
^ ≤^ ≤
2 , if 0 ( ) 0, if 0 1 4 , if > 1
x x f x x x x
^ <
2, if 1 ( ) 2 , if 1 1 2, if 1
x f x x x x
^ −^ ≤ −
17. Find the relationship between a and b so that the function f defined by 1, if 3 ( ) 3, if 3
ax x f x bx x
^ +^ ≤
+^ >
is continuous at x = 3.
18. For what value of λ is the function defined by
( 2 2 ), if 0 ( ) 4 1, if 0
x x x f x x x
λ − ≤ = +^ > continuous at x = 0? What about continuity at x = 1?
19. Show that the function defined by g ( x ) = x – [ x ] is discontinuous at all integral points. Here [ x ] denotes the greatest integer less than or equal to x. 20. Is the function defined by f ( x ) = x^2 – sin x + 5 continuous at x = π? 21. Discuss the continuity of the following functions: (a) f ( x ) = sin x + cos x (b) f ( x ) = sin x – cos x (c) f ( x ) = sin x. cos x 22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions. 23. Find all points of discontinuity of f , where sin , if 0 ( ) 1, if 0
x x f x x x x
24. Determine if f defined by
(^2) sin 1 , if 0 ( ) 0, if 0
x x f x (^) x x
is a continuous function?
CONTINUITY AND DIFFERENTIABILITY 119
f ( x ) xn^ sin x cos x tan x
f ′( x ) nxn^ – 1^ cos x – sin x sec^2 x
provided this limit exists. Derivative of f at c is denoted by f ′( c ) or (^ ( )) | c
d f x dx
. The
function defined by
0
( ) lim^ (^ )^ ( ) h
f x f^ x^ h^ f^ x → h
′ =^ +^ −
wherever the limit exists is defined to be the derivative of f. The derivative of f is
denoted by f ′ ( x ) or (^ ( ))
d f x dx
or if y = f ( x ) by
dy dx
or y ′. The process of finding
derivative of a function is called differentiation. We also use the phrase differentiate f ( x ) with respect to x to mean find f ′( x ).
The following rules were established as a part of algebra of derivatives: (1) ( u ± v )′ = u ′ ± v ′ (2) ( uv )′ = u ′ v + uv ′ (Leibnitz or product rule)
(3)^ u^ u v^ 2 uv v v
^ ′^ ′^ −^ ′
, wherever v ≠ 0 (Quotient rule).
The following table gives a list of derivatives of certain standard functions: Table 5.
Whenever we defined derivative, we had put a caution provided the limit exists. Now the natural question is; what if it doesn’t? The question is quite pertinent and so is
its answer. If 0
lim h
f c h f c → h
- − (^) does not exist, we say that f is not differentiable at c.
In other words, we say that a function f is differentiable at a point c in its domain if both
0 –
lim h
f c h f c → h
and 0
lim h
f c h f c →+ h
are finite and equal. A function is said
to be differentiable in an interval [ a , b ] if it is differentiable at every point of [ a , b ]. As in case of continuity, at the end points a and b , we take the right hand limit and left hand limit, which are nothing but left hand derivative and right hand derivative of the function at a and b respectively. Similarly, a function is said to be differentiable in an interval ( a , b ) if it is differentiable at every point of ( a , b ).
(^120) MATHEMATICS
Theorem 3 If a function f is differentiable at a point c , then it is also continuous at that point.
Proof Since f is differentiable at c , we have
( ) ( ) lim ( ) x c
f x f c f c → x c
But for x ≠ c , we have
f ( x ) – f ( c ) =
f x f c x c x c
Therefore lim [ x → c^ f^ ( ) x^^ −^ f c ( )]=
lim. ( ) x c
f x f c x c → x c
or lim [ x → c^^ f^ ( )] x^^ −^ lim [ x → c^ f c ( )]=
lim. lim [( )] x c x c
f x f c x c → (^) x c →
= f ′( c ). 0 = 0
or lim ( ) x c
f x →
= f ( c )
Hence f is continuous at x = c.
Corollary 1 Every differentiable function is continuous.
We remark that the converse of the above statement is not true. Indeed we have seen that the function defined by f ( x ) = | x | is a continuous function. Consider the left hand limit
0 –
lim 1 h
f h f h → h h
The right hand limit
0
lim 1 h
f h f h →+ h h
Since the above left and right hand limits at 0 are not equal, (^0)
lim h
f h f → h
does not exist and hence f is not differentiable at 0. Thus f is not a differentiable function.
5.3.1 Derivatives of composite functions
To study derivative of composite functions, we start with an illustrative example. Say, we want to find the derivative of f , where
f ( x ) = (2 x + 1)^3
(^122) MATHEMATICS
Put t = u ( x ) = x^2. Observe that cos
dv t dt
= and 2
dt x dx
= exist. Hence, by chain rule
df dx
= cos^2
dv dt t x dt dx
It is normal practice to express the final result only in terms of x. Thus
df dx
= (^) cos t ⋅ 2 x = 2 x cos x^2
EXERCISE 5.
Differentiate the functions with respect to x in Exercises 1 to 8.
1. sin ( x^2 + 5) 2. cos (sin x ) 3. sin ( ax + b ) 4. sec (tan ( (^) x )) 5.
sin ( ) cos ( )
ax b cx d
(^3). sin (^2) ( x (^5) )
7. (^) 2 cot ( (^) x^2 ) 8. (^) cos ( (^) x ) 9. Prove that the function f given by f ( x ) = | x – 1 |, x ∈ R is not differentiable at x = 1. 10. Prove that the greatest integer function defined by f ( x ) = [ x ], 0 < x < 3 is not differentiable at x = 1 and x = 2.
5.3.2 Derivatives of implicit functions Until now we have been differentiating various functions given in the form y = f ( x ). But it is not necessary that functions are always expressed in this form. For example, consider one of the following relationships between x and y :
x – y – π = 0 x + sin xy – y = 0 In the first case, we can solve for y and rewrite the relationship as y = x – π. In the second case, it does not seem that there is an easy way to solve for y. Nevertheless, there is no doubt about the dependence of y on x in either of the cases. When a relationship between x and y is expressed in a way that it is easy to solve for y and write y = f ( x ), we say that y is given as an explicit function of x. In the latter case it
CONTINUITY AND DIFFERENTIABILITY 123
is implicit that y is a function of x and we say that the relationship of the second type, above, gives function implicitly. In this subsection, we learn to differentiate implicit functions.
Example 22 Find
dy dx
if x – y = π.
Solution One way is to solve for y and rewrite the above as y = x – π
But then
dy dx
Alternatively , directly differentiating the relationship w.r.t., x , we have
d (^) ( x y ) dx
− (^) = d dx
π
Recall that
d dx
π means to differentiate the constant function taking value π
everywhere w.r.t., x. Thus
( ) ( )
d d x y dx dx
which implies that
dy dx
dx dx
Example 23 Find dy dx
, if y + sin y = cos x.
Solution We differentiate the relationship directly with respect to x , i.e.,
(sin )
dy d y dx dx
d x dx which implies using chain rule
cos
dy dy y dx dx
This gives
dy dx
sin 1 cos
x y
where y ≠ (2 n + 1) π
