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RATE OF CHANGE OF QUANTITIES: Ufa quanuty y varies with another quantity v, satisfying some rule v= f(x). Then or y= f(a) represents the nite of change of y with respect tox, dy , . . . at X= x, or (xy) represent the rate of change of y with respect to val v= xy ay © Iftwo variables vand_y varying with respect to another variable ¢, Le x= f() and y= g(t), then by . dy (dy ay dy chain rule — = #). “) es) dy \dr dt dt «The tenn nite of change will mean the instantaneous rate of change unless stated otherwise. * Marginal cost is the mte of change of total cost with respect to the output. * Marginal revenue is the nite of change of total revenue with respect to the output. ILLUSTRATION 1: WW xand y are the sides of wo squares such that y= .x—.1?. Find the rate of change of area of Second Square with respect to the first square. Explanation- Let 4, and ly denotes the areas of the First Square and Second Square respectively 4,=x? and Ay or Ay =(x- oS =%x-x7)1-2n) ILLUSTRATION 2: A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool ¢ seconds after the pool has been plugged off to drain and Z = 200(10-1)?. How fast is the water running out at the end of 5 seconds? Explanation- L = 200(10-1)? 4 ai —4100(10—1) dt (4) at 1=5 is equals to — 2000 dt So water is running out at the rate of 2000 litres/sceond RELATED RATES: Rate of change of one quantity involved is required corresponding to the given rate of change of another quantity. Example- Rate of change of volume of a spherical balloon is required when the rate of change of its radius is given. Scanned with CamScanner low fast is the volume of ec of lOen/sec. cube Example- An edge of a variable cube is increasing at the 2 when the edge Solution: Let xis the length of the edge of cube and I’ is the volume at any time f dh * and =10em/see dt Vex Example- Find angle @. in the first quadrant, which increases twice as fast as its sine. d Solution- dé =2 cf (sin 0) = 2cos@ do dt dt dt SOLUTION OF RATIONAL ALGEBRAIC INEQUATIONS The following results are very useful while solving rational algebraic inequations * ab>0>(a>0 or b>O)or(a(a>0 or b0) «© ab<0,a<0>h)<0 * ab>0,a>0>b>0 PO) >0 PO) 20 PO) <0 and * If P(x) and Q(x) are polynomials, then ineqautions , , Ox) Q(x) Ox) A(x) known as rational algebraic inequalities. ALGORITHM TO SOLVE RATIONAL ALGEBRAIC INEQUATIONS * Factorise P(x) and Q(x) into linear factors. * Make coefficients of x positive in all factors. + Equate all the factors to zero and find the corresponding value of x, These values are known as critical points. * Plot the critical points on the number line. * —n Critical points will divide the number line in 1+1 regions. * In the right most region, the expression will be positive and other regions altematively negative and positive. So mark positive sign in the right most region and then mar alternatively negative and positive signs in the remaining regions. © Obtain the solution set of the given inequation selecting the appropriate region. Scanned with CamScanner INCREASING AND DECREASING FUNCTIONS 7 : ; ora on increasing or decreas * By the use of differentiation, we can find out whether a function is increasing * Assume y=? and plot it in the plane. * As move from left to right - 0), the graph is decreasing. * As move from left to right (0,0), the graph is increasing. Vales left to ordain Values right to onain y [fwee ra 2 4 o[ =e 3 > a T “2 a 2 a - t oa) = 1 heignt of [od 7 1 1 f ereporn [> 7 2 a. _| sx. ae 7 = ee ener eet eee ee a o 2 rn as weave from ef fo Fight, the a6 we move fon lef raiaglM the elit of the graph decreases ¥ heigl af the araph increases INITION I(without derivatives) : Let J be an interval contained in the domain of a real valued function Then f is said to be * Strictly increasing on / if x, f(y) < f(x) .x.0, EF * Strictly increasing on /if x, >x,€/=> f(x) > Sx) 5%) ET * Swictly decreasing on / if x, > x, €/ => f(x) f(x%)> f(x) ,y,1 €/ * Constanton Jif f(x) =c forall x7, where cis a constant. Y fe) + ¥’ “I oma! x x Y Yy Strictly Increasing function Strictly Decreasing function Neither Increasing nor (i) rary) Decreasing function Gili) DEFINITION 2(without derivatives): Let J be an interval contained in the domain of a real valued function £ - Then / is said to be * Increasing on Jif x, f(x) < f(x) forall x,,x, €] * Increasing on Jif x, > x, in => f(x,)2 f(x;) forall x,x, €/ * Decreasing on / if x, > x, inI => f(x)< f(x,) forall x,x,e1 * Decreasing on / if x, f(x,)2 f(x) forall XX, ET ILLUSTRATION 1- (x) = e"is an increasing function Assume x,,x, € D, Let x, > 4, > f(x,)> f(x) ase" >e" So /(x)=e'is an increasing function Scanned with CamScanner ILLUSTRATION 2. /(.) s dee nevion tion DEFINITION 3 (By use of derivatives)- Let f(x) be a function which is di rentiable in (a,b). © s(x) * f(x) is dee f(x) MONOTONIC FUNCTION I Bin [ab] if f(x) <0 fora ye (a,b) ction in [a,b] if s’(Q) =0 forall xe (a,b) nereasing or decrea A real valued function is called monotonic if it cit 1 the give NECESSARY CONDITION FOR MONTONOCITY: Let f(x) be a function which is continuous in [ob] ‘crentiable in(a,b). Then, and * Wf fod is increasing on ( .b) vthen f"(x) >0 for all x € (a,b) e If f(x) is decreasing on (a,b) , then f(x) <0 foral ve (u,b) SUFFICEIENT CONDITION FOR MONTONOCITY- Let f(x) bea function wh h is continuous and differentiable in (a,b). Then, xe(a,b), then (x) is increasing on (a,6) XE (a,b) , then f(x) is decreasing on (a,b) © I f(x)>0 fo © If f(x) <0 fo J(x) = dx? = 6x? - 72x + 30 or S'(x) = 12% - 12x-72 = 12x? -x- 6) = 12(v-3) +2) Therefore, /’(x) = 0 gives x = —- 2, 3. The n f points x = - 2 and x =3 divides the real line into = > 3 te tluee disjoint intervals. namely, (2,2), (-2. 3) Fig 6.4 Examplel- aud (3,2). Interval Sign of f’(x) Nature of function / OO>9 Sis increasing OM 0 Sis increasing, Scanned with CamScanner wees xK “UONDI[JUL JO luLod payjes st jurod sty f “wUNTUTU (250) 10 PUILXELE [PIO[ Jo 1u1oOd ary sa9y}oU sto UaY ‘> YTNoMp seseoroUTX sv UBIs ay) saBurYa 10U saop (x),f J] * x 2 eurajur fe99] Jo juyod pac Ayuyepyssayp wou jo yuiod ewuyxce [929] Jo yujod puc Aunqepussayyp wou yo juyod etupreas pe90] . Jo uo SLD Udy) £970 1Y8Lt pue asoj> Atuaiayyns yurod Atsas 1e Q <(x),f puv ‘ojo yo] ay] OF asoya Aquatayyyns quod A1saa wg > (4), f Ju tortoy8nomp sosvatoulx sv aantsod 0] aanedau woy udis soBuey (x), f jie “BUNIXCW [Ld0] Jo jurod ay) pay[es S19 Vat) ‘9J0 ISU pur dsojo Atuatotysns ywIod Asaae ie Q>(x),f pure “240 YO] OU O1 asoja KjuaIOYJNS lutod Kana w Q< (x) f yi ‘at‘oySnomp saseosoury sv Sanesau 0} sanisod wor UsIs soueya (x) f JT» way 72 21d festa ay ye. snonunuos si («)/ pur y pearmu uado ay) uo pauyap uoNouny est (x) f 197 -LSUL SALLVARIIG LSU = Nae apqrnuaiayyip you st (x) f 10 9 =(9),f soya uayy 62 = ye BLUIUIL [90] JO BULIXELL |LI0] eB sey (x) f J] wuiod Aue aq 7 39 asoddng + J [earqur uado ue ut pauyyop uorouny v aq (X)/ 197 -IWIOFILL (y4+2*y-2) [earaqut Py UL apqenuasayyp se(x)f yey yons Q0 Close to 1 (° the left (say 0.9 ete.) <0 to the right (say — 0.9 ete.) <0 Close to 1 ( the left (say - 1.1 ete.) >0 SECOND DERIVATIVE TEST = Let f(r) be a function defined on the interval J andee/. Let f(x) be \Wice differentiable ate, Then ¢ is point of local maxima if f"(c)=0 and f"(c) <0. The value of f(c) is called local maximum eox value. * x=c is point of local minima if /'(c)=0 and f"(c)>0. The value of f(c) is called local minimum value. + Thetest failsif /'(c)=0 and f"(c)=0 \ Twice differentiable means second order derivative ABSOLUTE MAXIMA OR MINIMA- Let f(x) be a continuous function on an interval J e (a,b). Then f has absolute maximum value and /f attains it at least once in/. Also f has absolute minimum value and f attains it at least once in J . Working Rule «Find all critical points of / on the given interval. * Take the end points of interval. * Atal these points calculate the value of «Identify the maximum and minimum value of f calculated * The greatest value of £ is absolute maxima * Thesmallest value of f is absolute minima Y | ir@ ive ire. Hic “Ola : : Ax ¥, Scanned with CamScanner 
