unit 1 differential calculus - IGNOU

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Mathematics-II 18 Two special cases of the above definition are important. (i) Given any function f, if k = 0, the function kf turns out to be the zero function. That is, kf = 0. (ii) If k = − 1, the function kf is called the negative of f and is denoted simply by – f. Absolute Value Function (or Modulus Function) of a Given Function Let f be a function with domain D. The absolute value of function f denoted by | f | and read as mod f is defined by setting (| f |) (x) = | f (x) | for all x ∈ D. Since | f (x) | = f (x), if f (x) ≥ 0, thus | f | have the same graph for those values of x for which f (x) ≥ 0. Now let us consider those values of x for which f (x) < 0. Here | f (x) | = − f (x). Therefore, the graphs of f and | f | are reflections of each other w. r. t. the x-axis for those values of x for which f (x) < 0. As an example, consider the graph in Figure 1.9(a). The portion of the graph below the x-axis, that is, the portion for which f (x) < 0 has been shown as dotted. To draw the graph of | f | we retain the undotted portion in Figure 1.9(a) as it is, and replace the dotted portion by its reflection w. r. t. the x-axis (Figure 1.9(b)). y 0 x (a) (b) Figure 1.9 Sum, Difference, Product and Quotient of Two Functions If we are given two functions with a common domain, we can form several new functions by applying the four fundamental operations of addition, subtraction, multiplication and division on them. (i) Define a function δ on D by setting δ (x) = f (x) + g (x). The functions is called the sum of the functions f and g, and is denoted by f + g. Thus, (f + g) (x) = f (x) + g (x). (ii) Define a function d on D by setting d (x) = f (x) – g (x). The function d is the function obtained by subtracting g from f, and is denoted by f – g. Thus, for all x ∈ D, (f − g) (x) = f (x) – g (x). (iii) Define a function p on D by setting p (x) = f (x) g (x). The function p is called the product of the functions f and g, and is denoted by fg. Thus, for all x ∈ D, (fg) (x) = f (x) g (x). y 0 x
Remark (iv) Define a function q on D by setting f ( x) q ( x) = . The function p is g ( x) provided g (x) ≠ 0 for any x ∈ D. The function q is called the quotient f f f ( x) of f by g and is denoted by . Thus, ( x) = , g g g ( x) (g (x) ≠ 0 for any x ∈ D). In case g (x) = 0 for some x ∈ D, we can consider the set, say D′ of all those f f f ( x) values of x for which g (x) ≠ 0 and define on D′ by setting ( x) = g g g ( x) for all x ∈ D′. Example 1.4 Consider the functions f : x → x 2 and g : x → x 3 . Find the functions which are defined as (i) f + g, (ii) f – g, (iii) fg, (iv) (2f + 3g), and (v) Solution f . g (i) (f + g) (x) = x 2 + x 3 (ii) (f – g) (x) = x 2 – x 3 (iii) (fg) (x) = x 5 (iv) (2f + 3g) (x) = 2f (x) + 3g (x) = 2f (x) + 3g (x) = 2x 2 + 3x 3 . (v) Now, g (x) = 0 ⇔ x 2 = 0 ⇔ x = 0. Therefore, in order to define the f function , we shall consider only non-zero values of x. If x ≠ 0, g 2 f ( x) x 1 f f 1 = = . Therefore, is the function : x → , whenever g ( x) 3 x x g g x x ≠ 0. 1.3.6 Composition of Functions We shall now describe a method of combining two functions which is somewhat different from the ones studied so far. Uptil now, we have considered functions with the same domain. We shall now consider a pair of functions such that the co-domain of one is the domain of the other. Let f : X → Y and g : Y → Z be two functions. We define a function h : X → Z by setting h (x) = g (f (x)). Differential Calculus 19
Page 1 and 2: UNIT 1 DIFFERENTIAL CALCULUS Struct
Page 3 and 4: Definition 2 Let S be a non-empty s
Page 5 and 6: 1 1 (ii) = , if x ≠ 0 x | x| (iii
Page 7 and 8: The role that the graph of a functi
Page 9 and 10: The Greatest Integer Function Figur
Page 11 and 12: Consider the function f : N → E d
Page 13: (iii) f : R → R defined by f (x)
Page 17 and 18: ⎛ x 3 ⎞ ⎛ x 3 ⎞ fog (x) = f
Page 19 and 20: 0 Figure 1.13 (ii) The function g d
Page 21 and 22: Solution Look at Tables 1.1 and 1.2
Page 23 and 24: if, and only if, for each ε > 0 th
Page 25 and 26: (vi) lim x →a [ lim f ( x)] ⎡ f
Page 27 and 28: SAQ 4 lim = x →0 Evaluate the lim
Page 29 and 30: Remark If the interval is closed, t
Page 31 and 32: Example 1.20 Solution Draw the grap
Page 33 and 34: 1.6.1 Derivative of a Function We n
Page 35 and 36: Remark f ( 0 + h) − f ( 0) h −
Page 37 and 38: (ii) (iii) d dg ( x) df ( x) ( f (
Page 39 and 40: 1.6.3 Derivative of a Composite Fun
Page 41 and 42: (iii) (x 2 + x + 6) sin x (iv) (x +
Page 43 and 44: dy x Therefore, = y log e a = a log
Page 45 and 46: facilitates differentiation. Second
Page 47 and 48: (c) Differentiate the following (i)
Page 49 and 50: SAQ 5 SAQ 6 SAQ 7 SAQ 8 (viii) Limi
Page 51: MATHEMATICS-II The knowledge and co

unit 1 differential calculus - IGNOU

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