Chapter 5.1 Overview 5.1.1 Continuity of a function at a point Let fbe a real function on a subset of the real numbers and let cbe a point in the domain of f. Then fis continuous at cif lim f() xf= () c xc→ More elaborately, if the left hand limit, right hand limit and the value of the function at x = c exist and are equal to each other, i.e., lim xc ( ) fx f ( ) c lim xc f ( ) x then f is said to be continuous at x = c. 5.1.2 Continuity in an interval (i) fis said to be continuous in an open interval (a, b) if it is continuous at every point in this interval. (ii) fis said to be continuous in the closed interval [a, b] if • fis continuous in (a, b) • lim x→a+ f(x) = f(a) • lim f(x) = f(b)x→b– 5.1.3 Geometrical meaning of continuity (i) Function f will be continuous at x = c if there is no break in the graph of the function at the point (cfc .,()) (ii) In an interval, function is said to be continuous if there is no break in the graph of the function in the entire interval. 5.1.4 Discontinuity The function f will be discontinuous at x = a in any of the following cases : (i) lim f (x) and lim f (x) exist but are not equal.x→a− x→a+ (ii) lim f (x) and lim f (x) exist and are equal but not equal to f (a).x→a− x→a+ (iii) f (a) is not defined. 5.1.5 Continuity of some of the common functions Function f (x) Interval in which f is continuous 1. The constant function, i.e. f (x) = c 2. The identity function, i.e. f (x) = x R 3. The polynomial function, i.e. f (x)= a0 xn + a1 xn–1 + ... + ax + a n–1 n 4. | x – a | 5. x–n, n is a positive integer 6. p (x) / q (x), where p (x) and q (x) are polynomials in x 7. sin x, cos x (– ∞ , ∞ ) (– ∞ , ∞ ) – {0} R – { x : q (x) = 0} R 8. tan x, sec x R– { (2 n + 1) π 2 : n ∈ Z} 9. cot x, cosec x R– { (nπ : n ∈ Z} 10. ex R 11. log x (0, ∞) 12. The inverse trigonometric functions, In their respective i.e., sin–1 x, cos–1 x etc. domains 5.1.6 Continuity of composite functions Let f and g be real valued functions such that (fog) is defined at a. If g is continuous at a and f is continuous at g (a), then (fog) is continuous at a. 5.1.7 Differentiability f (xh) x+−f () The function defined by f ′ (x) = lim , wherever the limit exists, ish→0 h defined to be the derivative of f at x. In other words, we say that a function f is f (ch) c+− f () differentiable at a point c in its domain if both lim − , called left handh→0 h f (ch) c+− f () derivative, denoted by Lf ′ (c), and lim + , called right hand derivative,h→0 h denoted by R f ′ (c), are finite and equal. (i) The function y = f (x) is said to be differentiable in an open interval (a, b) if it is differentiable at every point of (a, b) (ii) The function y = f (x) is said to be differentiable in the closed interval [a, b] if R f ′ (a) and L f ′ (b) exist and f ′ (x) exists for every point of (a, b). (iii) Every differentiable function is continuous, but the converse is not true 5.1.8 Algebra of derivatives If u, v are functions of x, then du( ±v) dudv d dv du (i) =± (ii) () =u +vdx dxdx dxuv dx dx du dv v −u⎛⎞(iii) du =dx dx ⎜⎟2dx ⎝⎠vv 5.1.9 Chain rule is a rule to differentiate composition of functions. Let f = vou. If dt dv df dvdt t = u (x) and both and exist then = . dxdtdx dtdx 5.1.10 Following are some of the standard derivatives (in appropriate domains) 1. –1(sin d dx )x = 1 1− 2x 2. –1(cos )d xdx = 2 1 1 x − − 3. –1 2 1(tan ) 1 = + d x dx x 4. –1 2 1(cot ) 1 d x dx x − = + 5. –1(sec )d xdx = 2 1 xx , 1− x 1> 6. –1(cosec )d x dx = 2 1 xx − , 1− x 1> 5.1.11 Exponential and logarithmic functions (i) The exponential function with positive base b > 1 is the function y = f (x) = bx. Its domain is R, the set of all real numbers and range is the set of all positive real numbers. Exponential function with base 10 is called the common exponential function and with base e is called the natural exponential function. (ii) Let b > 1 be a real number. Then we say logarithm of a to base b is x if bx=a, Logarithm of a to the base b is denoted by logba. If the base b = 10, we say it is common logarithm and if b = e, then we say it is natural logarithms. logx denotes the logarithm function to base e. The domain of logarithm function is R+, the set of all positive real numbers and the range is the set of all real numbers. (iii) The properties of logarithmic function to any base b > 1 are listed below: 1. logb (xy) = logbx + logby ⎛⎞x 2. logb ⎜⎟= logbx – logby3.logb xn = n logb x y⎝⎠ logcx 4. log b =x , where c > 1log cb 1 =5. logbx log xb 6. logbb = 1 and logb 1 = 0 d x x(iv) The derivative of ex w.r.t., x is exdx ()e . The derivative of logx, i.e. e 1 d 1 xw.r.t., x is ; i.e. (log ) . xdx x 5.1.12 Logarithmic differentiation is a powerful technique to differentiate functions of the form f (x) = (u (x))v(x), where both f and u need to be positive functions for this technique to make sense. 5.1.13 Differentiation of a function with respect to another function Let u = f (x) and v = g (x) be two functions of x, then to find derivative of f (x) w.r.t. du to g (x), i.e., to find , we use the formuladv du du dx= dv dv . dx 5.1.14 Second order derivative 2ddy dy 2 is called the second order derivative of y w.r.t. x. It is denoted by y′′ ordx dx dx y2 , if y = f (x). 5.1.15 Rolle’s Theorem Let f : [a, b] R be continuous on [a, b] and differentiable on (a, b), such that f (a) = f (b), where a and b are some real numbers. Then there exists at least one point c in (a, b) such that f ′ (c) = 0. Geometrically Rolle’s theorem ensures that there is at least one point on the curve y= f(x) at which tangent is parallel to x-axis (abscissa of the point lying in (a, b)). 5.1.16 Mean Value Theorem (Lagrange) Let f: [a, b] R be a continuous function on [a, b] and differentiable on (a, b). Then f()b () fa there exists at least one point cin (a, b) such that f′ (c) = .ba Geometrically, Mean Value Theorem states that there exists at least one point cin (a, b) such that the tangent at the point (c, f(c)) is parallel to the secant joining the points (a, f(a) and (b, f(b)). 5.2 Solved Examples Short Answer (S.A.) Example 1 Find the value of the constant kso that the function fdefined below is ⎧ ⎪⎪1– cos4 xfxcontinuous at x= 0, where () =⎨ 2, x≠0.8x⎪ ⎪k, x=0⎩ Solution It is given that the function fis continuous at x= 0. Therefore, lim f(x) = f(0)x→0 1– cos4 x⇒ lim 2 = k x→0 8x 2sin 2 2x⇒ lim 2 = k x→08x ⎛ sin 2 x⎞2 ⇒ lim = k⎜⎟ x→0 ⎝ 2x ⎠ ⇒ k= 1 Thus, fis continuous at x= 0 if k= 1. Example 2 Discuss the continuity of the function f(x) = sin x. cos x. Solution Since sin xand cos xare continuous functions and product of two continuous function is a continuous function, therefore f(x) = sin x. cos xis a continuous function. ⎧ x3 + x2 –16 x+ 20 ⎪ 2, x≠ 2 Example 3 If fx() =⎨ x(– 2) is continuous at x= 2, find ⎪⎩k , x=2 the value of k. Solution Given f(2) = k. x3 + x2 –16 x+ 20 Now, lim fx= lim fx = lim () () x 2x→2– x→2+→ (– 2) 2x (x 5)( x– 2) 2 = lim 2 lim( x 5) 7 2 x 2(–2) xx As fis continuous at x= 2, we have lim f() x= f(2) x→2 ⇒ k= 7. Example 4 Show that the function fdefined by ⎧ 1⎪xsin , x≠ 0fx() =⎨ x ⎪⎩ 0, x=0 is continuous at x= 0. Solution Left hand limit at x= 0 is given by fx lim lim () =xsin 1 = 0 [since, –1 < sin 1 < 1]–– xxx→0 x→0 fxSimilarly lim ( ) lim xsin 1 0 . Moreover f(0) = 0.x 0 x 0 x fx (0) Thus lim 0– f() x lim () f . Hence fis continuous at x= 0 xx 0 1Example 5 Given f(x) = . Find the points of discontinuity of the compositex–1 function y = f[f(x)]. 1Solution We know that f(x) = is discontinuous at x= 1 x–1 Now, for x 1, 1 x –1 f (f (x)) = f 1 = 1 2– x ,–1x–1 x –1 which is discontinuous at x = 2. Hence, the points of discontinuity are x = 1 and x = 2. Example 6 Let f(x) = x x , for all x ∈ R. Discuss the derivability of f(x) at x = 0 ⎧ x2,if x ≥ 0fxSolution We may rewrite f as () = ⎨⎪ ⎪−x2,if x <0⎩ f (0 + h)– f (0) –h2 –0 Now Lf ′ (0) = lim = lim = lim − h = 0 –– −h→0 hh→0 hh→0 f (0 + h)– f (0) h2 –0 Now Rf ′ (0) = lim = lim = lim h = 0 + +−h→0 hh→0 hh→0 Since the left hand derivative and right hand derivative both are equal, hence f is differentiable at x = 0. Example 7 Differentiate tan xw.r.t. x Solution Let y = tan x . Using chain rule, we have dy 1 d . (tan x)dx dx2tan x d1 .sec 2 x ( x)= dx2tan x 12 1(sec x)= 2 x2tan x dyExample 8 If y = tan(x + y), find dx . Solution Given y = tan (x + y). differentiating both sides w.r.t. x, we have dy 2dsec ( xy)(xy)dx dx dy = sec2 (x + y) 1 dx dyor [1 – sec2 (x + y] dx = sec2 (x + y) dy sec 2 (xy)Therefore, 2 = – cosec2 (x + y).dx 1sec (xy) Example 9 If ex + ey = ex+y, prove that dy =−eyx . − dx Solution Given that ex + ey = ex+y. Differentiating both sides w.r.t. x, we have dy dy = ex+y 1ex + ey dxdx dyor (ey – ex+y) = ex+y – ex,dx xyx xyxdy e – e eee – ewhich implies that yxy yxy yx .dxee e ee dy ⎛3xx− 3 ⎞ 11