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1 MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give n exmple of function with infinitely mny verticl symptotes. Solution. The function y = (x )(x 2) (x n), with domin of definition R \ {, 2,..., n}, hs exctly n verticl symptotes, nmely t x =,..., n. In generl, given n distinct numbers,..., n, the function y = (x )(x 2) (x n), with domin of definition R \ {,..., n }, hs exctly n verticl symptotes, nmely t x =,..., n. The function y = tn x, with domin of definition R \ {..., 3π 2, π 2, π 2, 3π 2,... }, hs infinitely mny verticl symptotes, exctly t x {..., 3π 2, π 2, π 2, 3π 2,... }, which is where cos x = 0. Exercise 2. Let f be function which is differentible everywhere. Suppose tht f (x) > for ll x. Show tht x f(x) =. Solution. Let x < x 2 be rel numbers. By the Men Vlue Theorem, there exists some c (x, x 2 ) such tht f (c) = f(x2) f(x) x 2 x, nd since f (c) > we hve f(x 2 ) f(x ) > x 2 x. In prticulr, substituting x 2 = x nd x = 0, we hve f(x) > x + f(0) for ll x > 0. Thus, since x (x + f(0)) =, we hve x f(x) =. An lternte solution is to use the Fundmentl Theorem of Clculus. We hve f(x) f(0) ( ) = x where we used FTC in the equlity mrked ( ) bove. 0 f (t) dt > x 0 dt = x Be creful bout using indefinite integrls. It is not necessrily true tht f(x) > x for ll x; it is only true for ll x greter thn some number. For exmple, consider the cse f(x) = 2x; then f (x) = 2 so f (x) > for ll x. But f(x) > x only if x > 0. Exercise 3. Grph the function f(x) = x 3 + 6x 2 + 9x. Indicte domin, criticl points, inflection points, regions where the grph is incresing/decresing, x- intercepts nd y-intercepts, regions of concvity (up or down), locl mxim nd minim, ny symptotes nd behvior t infinity. Solution. The domin is R. A point (x, f(x)) on the grph of the function is criticl point if f (x) = 0 or is not defined; f(x) is differentible everywhere, so the only wy point cn be criticl point is if f (x) = 0. We hve f (x) = 3x 2 + 2x + 9 = 3(x + )(x + 3)

2 2 FRI, OCT 25, 203 so the only criticl points of f re (, f( )) = (, 4) nd ( 3, f( 3)) = ( 3, 0). A point (x, f(x)) on the grph of the function is n inflection point if f (x) = 0 or is not defined; gin, the only wy point cn be n inflection point is if f (x) = 0. We hve f (x) = 6x + 2 so the only criticl point of f is ( 2, f( 2)) = ( 2). We cn fctor f(x) = x(x + 3) 2, so the x-intercepts re x = 0, 3. The y-intercept is f(0) = 0. The function f is incresing (resp. decresing) in the intervl I if f(x ) < f(x 2 ) (resp. f(x ) > f(x 2 )) for ll x, x 2 I such tht x < x 2 (this is the definition on pge 9). This is equivlent to sying tht f (x) > 0 (or f (x) < 0) for ll x in the intervl. Since f (x) > 0 on (, 3) nd (, ) nd f (x) < 0 on ( 3, ), f is incresing on (, 3), decresing on ( 3, ), nd incresing on (, ). A function f is concve up (resp. down) on the intervl I if f (x) > 0 (resp. f (x) < 0) for ll x I. Since f (x) = 6(x + 2), f (x) > 0 on the intervl ( 2, ) nd f (x) < 0 on the intervl (, 2). So f is concve up on ( 2, ) nd concve down on (, 2). A function which is differentible on ll of R hs locl mximum (resp. minimum) t x = if nd only if f () = 0 nd f () < 0 (resp. f () > 0). We hve f ( ) = 0 nd f ( ) = 6 > 0, so f hs locl minimum t x = ; we hve f ( 3) = 0 nd f ( 3) = 6 < 0, so f hs locl mximum t x = 3. There re no verticl symptotes becuse the function is continuous on the entire rel line. Suppose there is slnt or horizontl symptote; then there exist rel numbers m, b such tht either (f(x) (mx + b)) = 0 or (f(x) (mx + b)) = 0. x x But f(x) (mx+b) is polynomil of degree 3, so we hve contrdiction (recll the exercise bout endpoint behvior of nonconstnt polynomils). So there re no slnt or horizontl symptotes. y ( 3, 0) x ( 2, 2) (, 4) Figure. Grph of f(x) = x 3 + 6x 2 + 9x Notice tht the function f(x) = x defined on R\{0} hs the property tht f (x) < 0 for ll x R\{0}, but it is not decresing on R \ {0} (since f( ) < f()). This hs to do with the fct tht f is not defined t x = 0.

3 MORE FUNCTION GRAPHING; OPTIMIZATION 3 Exercise 4. Find t 6 in three wys: (i) using methods lerned up to nd including the first midterm; (ii) by relizing the it s f (c) for some function f(t) nd some vlue c; (iii) using L Hospitl s Rule. Solution. (i) We hve (ii) Set f(t) = t. Then (iii) We hve t 6 = ( )( t + 4) = t + 4 = 8. t 6 = f(t) f(6) = f (6) = /2 = t /2 t 6 = t = 8. Exercise 5 (Section 4.7, #9). Find the point on the line y = 2x + 3 tht is closest to the origin. Solution. At the point (x, 2x + 3), the distnce to the origin is d(x) = (x 0) 2 + (2x + 3 0) 2 = x 2 + (2x + 3) 2 = 5x 2 + 2x + 9 by the Pythgoren theorem. The function d(x) is defined for ll x nd is differentible everywhere. We wnt to find globl minimum of d(x). To ese computtion, I mke the following (perhps unconventionl) rgument. Let p(x) = 5x 2 + 2x + 9. For ny two nonnegtive rel numbers, the condition x x 2 is equivlent to x x 2. Thus, for ny two rel numbers x, x 2, the condition d(x ) d(x 2 ) is equivlent to p(x ) p(x 2 ) (since d(x) = p(x)). Thus d() is globl minimum vlue of the function d(x) if nd only if p() is globl minimum vlue of the function p(x). Let s find the globl minimum vlue of p(x). For qudrtic polynomils x 2 + bx + c with > 0, the minimum vlue tkes plce t x = b 2 since tht s the vertex of the prbol. 2 So in our cse the minimum of p(x) occurs t x = = 6 5. Thus the desired point is ( 6 5, 2( 6 5 ) + 3) = ( 6 5, 3 5 ).3 Exercise 6 (Section 4.7, #2). Find the points on the ellipse 4x 2 + y 2 = 4 tht re frthest wy from the point (, 0). Solution. I m going to find the point on the upper hlf of the ellipse which is frthest from the point (, 0), then note tht the ellipse is symmetric with respect to the x-xis, so the reflection will be frthest, too. The eqution of the upper prt is given by y = 4 4x 2, with domin of definition [, ]. It is continuous on [, ] nd differentible on (, ). The distnce from the point (x, 4 4x 2 ) to (, 0) is given by d(x) = (x ) 2 + ( 4 4x 2 0) 2 = 3x 2 2x + 5 by the Pythgoren theorem. It is continuous on [, ] nd differentible on (, ). Set p(x) = 3x 2 2x + 5. By the rgument given in the solution to Exercise 5, the problem reduces to finding the mximum 2 Exercise: Prove this. 3 Note tht the line joining ( 6 5, 3 ) to the origin is perpendiculr to the line y = 2x + 3. This will lwys be the cse for 5 problems of the form find the point on the line l tht is closest to the point P.

4 4 FRI, OCT 25, 203 of the qudrtic p(x) on the intervl [, ]. If the qudrtic were defined over ll of R, then its globl mximum would occur t x = = 3, which is inside the intervl [, ], so this globl mximum of the qudrtic on the intervl [, ]. Thus the point ( 3 4, 4( 3 )2 ) = ( 3, ) is the point on the upper hlf of the ellipse which is frthest wy from (, 0). On the bottom hlf, the point ( 3, ) is frthest wy. So the desired points re ( 3, ) nd ( 3, ). Exercise 7 (Section 4.7, #24). Find the re of the lrgest rectngle tht cn be inscribed in the ellipse x 2 / 2 + y 2 /b 2 =. Solution. Assume without loss of generlity tht, b re positive. If b =, then the ellipse is ctully circle nd ny rectngle inscribed in circle of rdius is squre of side length 2, so it hs re 2 2. Now ssume b <. I m going to ssume without proof tht, for ny rectngle inscribed in n ellipse which is not circle, its sides re prllel to the xes of the ellipse. 4 Since the sides of our rectngles re prllel to the x nd y xes, its vertices re (x, y), ( x, y), ( x, y), (x, y) for some x [0, ] nd y = b x2. 2 Thus the re of the rectngle is A(x) = 2x 2y = 4bx x2 2 where A(x) is function on the domin [0, ]. It is continuous on [0, ] nd differentible on (0, ). We mximize A(x). Let p(x) = x 2 2 x 4. Then A(x) = 4b p(x). By the rgument given in the solution to Exercise 5, the problem reduces to finding the mximum of p(x) on the intervl [0, ]. Considering p(x) to be qudrtic polynomil in the vrible x 2, we hve tht it tkes mximum when x 2 = = , or when x = 2, which is inside the intervl [0, ]. For this vlue of x, we hve y = b (/ 2) 2 = b 2 2, so the rectngle hs re b 2 = 2b. 5 Exercise 8 (Section 4.7, #54). At which points on the curve y = + 40x 3 3x 5 does the tngent line hve the lrgest slope? Solution. Let f(x) = + 40x 3 3x 5. At (x, f(x)), the tngent line to the curve y = f(x) hs slope f (x) = 20x 2 5x 4. So our problem reduces to mximizing f (x). We hve f (x) = 240x 60x 3 = 60x(x 2)(x + 2), which is 0 only if x = 0, ±2. We hve f (x) = x 2 nd f (0) = 240 > 0, f (2) = 480 < 0, f ( 2) = 480 < 0, so 0 is locl minimum while ±2 re locl mxim. Since f (x) > 0 on the intervls (, 2) nd (0, 2) nd f (x) < 0 on the intervls ( 2, 0) nd (2, ), we hve tht f (x) is incresing on the intervls (, 2) nd (0, 2) nd f (x) is decresing on the intervls ( 2, 0) nd (2, ). Thus ( 2, f( 2)) = ( 2, 223) nd (2, f(2)) = (2, 225) re the globl mxim of f (x). Alterntively, you cn complete the squre in f (x). We hve f (x) = 5(x 4 8x 2 ). Completing the squre gives f (x) = 5(x 4 8x 2 + 6) = 5(x 2 4) , so f (x) tkes its mximum whenever 4 This is minor technicl point which is less relevnt to Mth A but is still necessry in complete solution: if rectngle is inscribed in n ellipse which is not circle, how do I know tht its sides re going to be prllel to the x nd y xes? (For exmple, if the ellipse is circle, then inscribed rectngles re squres, whose sides don t hve to be prllel to the xes.) I hope this following rgument will stisfy you. Assume without loss of generlity tht b < nd rectngle is inscribed in the ellipse such tht its sides ren t prllel to the xes of the ellipse. Apply the liner trnsformtion which shrinks everything long the x-xis by the fctor k = b. So the ellipse now becomes circle of rdius b. Under this trnsformtion, the rectngle tht ws inscribed in the ellipse becomes prllelogrm. The four sides of the originl rectngle hve slope m, m, m, m for some nonzero m, so the prllelogrm hs four sides whose slopes re mk, k m, mk, k. Since k >, djcent sides of the m prllelogrm ren t perpendiculr, nd the prllelogrm is not rectngle. Thus the prllelogrm hs n cute ngle, nd such prllelogrms cnnot be inscribed in circles, contrdiction. 5 Note tht this gives the sme nswer s bove for the specil cse b =.

5 MORE FUNCTION GRAPHING; OPTIMIZATION 5 (x 2 4) 2 is minimized, i.e. when x 2 4 = 0, or x = ±2. Thus our points re ( 2, f( 2)) = ( 2, 223) nd (2, f(2)) = (2, 225).

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