THE EXTREMA OF THE RATIONAL QUADRATIC FUNCTION f(x)(x +x+b)/(x +x+d) Lrry Lrson & Slly Keely Rtionl funtions nd their grphs re explored in most prelulus ourses. Mny websites provide intertive investigtions of the zeroes, vertil symptotes, nd horizontl symptote of these funtions. We present, we think, surprising reltionship between the extrem of the prtiulr rtionl funtion x + x + b ( x z1)( x z) nd the zeroes, z 1, z, z 3, z 4 of the qudrtis involved. We x + x + d ( x z3)( x z4) believe the importne of the horizontl symptote, y1, hs been overlooked in the literture. We provide method for finding the ext extrem without the use of lulus, one tht unexpetedly involves the rithmeti nd geometri mens of the expressions involving the zeroes. Let us first present the min theorem then illustrte with n exmple. The proof of the theorem follows with some dditionl results. We onlude with some suggestions for further explortion. Let THEOREM: x + x + b ( x z1)( x z) N( x) where the z i my be rel or omplex. x + x + d ( x z3)( x z4) D( x) Let (e,1) be the point of intersetion of f(x) with the horizontl symptote y1. Then d b e,, nd the extrem of f(x) our t (x,y) where ( 1)( ) (( 1),( ) ) (( 1),( ) ) ± (( 1),( ) ) (( ),( )) (( ),( )) x e ± e z e z e ± GM e z e z y 3 4 ± 3 4,, AM refers to the rithmeti men, nd GM refers to the geometri men. We illustrte this with n exmple. ( x 1)( x + 1) x 1 ( x )( x + 3) x + x 6 EXAMPLE 1: d b 6 + 1 (4)(6) e 5 We note tht f (5) (3)(8) 1. 0 1 Now by the theorem, we find tht the oordintes of the reltive extrem re, x e ± GM ( e z ),( e z ) 5 ± GM 4,6 5 ± 4 6 5 ± 4 nd, 1
AM ( e z 4 6 1),( e z) ± GM ( e z1),( e z) AM 4,6 ± GM 4,6 + ± 4 6 5 ± 4 y AM ( e z ),( e z ) ± GM ( e z ),( e z ) AM 3,8 ± GM 3,8 ± 3 8 ± 4 3+ 8 11 3 4 3 4 Tht is, 5 + 4, 5 + 4 11 + 4 nd 5 4, 5 4. No lulus required! 11 4 Note the ese of the lultions using the distnes from e to the zeroes. PROOF OF THEOREM: Let ( ) ( ) x + x + b x z1 x z N x x + x + d x z3 x z4 D x
The ritil vlues k of f(x) our where, f '( k) 0 D( k) N'( k) N( k) D'( k) 0 N( k) N'( k) D( k) D'( k) k + k + b k + k + k + d k + Cross multiplying nd simplifying gives, + + + + + + + + k k bk k b k k dk k d ( ) + ( ) (Rell: ) d b k k b d d b e d b k ek + e + e d b + b b ( k e) + e ( d b) b( ) ( k e) + + e ( k e) e + b + e ± + + ± ( )( ) k e e e b e e z1 e z Note if e + e + b < 0 there re no ritil vlues. N( k) N'( k) k + We find f( k) D( k) D'( k) k + whih implies, k + f( k) k + e ± ( e z )( e z ) + (Note: + ) e ± ( e z )( e z ) + 1 e+ e+ ( z1 z) e 3 4 ( e z1) + ( e z) ± ( e z1)( e z) ( e z3) + ( e z4) ± ( e z3)( e z4) (( 1),( ) ) ± (( 1),( ) ) (( ),( )) ± (( ),( ) 3 4 3 4 )
Let s look t n exmple where the grph of f hs no vertil symptotes, but still rosses the horizontl symptote. x + x 3 ( x 1)( x + 3) ( x 1)( x + 3) EXAMPLE : x + x + 1 x + x + 1 1 + i 3 1 i 3 x x d b 1 + 3 e 4. Now by the theorem, we find tht the oordintes of the reltive extrem re, 1 x e ± GM ( e z ),( e z ) 4 ± GM 3,7 4 ± 3 7 4 ± 1 nd, ( 1 ) ( ) ( 1 ) ( 1 ) AM ( e z ),( e z ) ± GM ( e z ),( e z ) 5 ± 1 y AM ( e z ),( e z ) ± GM ( e z ),( e z ) ± 1 Tht is, 4 + 1, 9 3 4 3 4 5 + 1 9 + 1 nd 4 1, 5 1. Agin, no lulus needed. 9 1 Cse where : d b If then e ± lim e e + e + b.. The ritil vlue of f(x), k, is then ( ) e ± b e b e + + e + e + e + b e e lim e e e b lim lim PROOF: ( ) e + e + e + b 1 + 1 + + (The se of the limit s e is left to the reder.)
Now, ( ) + ( ) + b ( ) + ( ) + d f( ) 4 4 + + b d z1 z ( ) z3 z4 ( ) (*) + + [ AM( z1, z) ] [ GM( z1, z) ] [ AM( z, z )] [ GM( z, z )] AM AM 3 4 3 4 b d + + GM GM * Interestingly this is (of the zeros z ) i 4b, the rtio of the disriminnts of N(x) nd D(x). 4d We lso note tht when, N(x) nd D(x) hve the sme xis of symmetry, nd f(x) is symmetri bout x. Showing tht following exmple: f( + x) f( x) we leve s n exerise. We present for observtion, the One further theorem is of interest. THEOREM: If 4b 4d, then f( e + x) 1. f( e x)
PROOF: We show tht N(e+x) D(e-x). e + x + e + x + b e x + e x + d 4ex + x + x + e e + b d 0 d b x(4 e + + ) + e( ) + b d 0 (Rell: e ) x d b b d 4d 4b+ ( ) + + 0 ( 4 b) ( 4 d) x( ) 0 Sine 4b 4d we hve x (0) 0 Similrly D(e+x) N(e-x), thus f( e + x). In prtiulr, fmx N( e+ x) D( e x) 1 D( e+ x) N( e x) f ( e x) 1 f. min EXAMPLE 3: d b e 3 x x ( x + 1)( x ) x + 7x + 10 ( x + )( x + 5) 1 7 ( x)( x). f( e + x) + + whih is the reiprol of 1 7 ( + x)( + x) By the min theorem, the reltive extrem re Note tht 1 1 4 + 7 4 7 4 + 7 4 7 4+ 7 fmin 4+ 7 3 + 7 4 + 7,.nd 4 + 7 f mx. f( e x) ( x)( x). ( )( x) 1 7 1 7 x 3 7 4 7,. 4 7 Lstly, we enourge the reder to investigte the following relted funtions: x + b 1. with horizontl symptote y0 x + x + d. with horizontl symptote y0 x + x + d x + bx + 3. with horizontl symptote y/d dx + ex + f We hve found mny interesting geometri nunes relted to these funtions worthy of further investigtion. We wish to thnk Russell Gordon of Whitmn College, Wll Wll, Wshington, for his geometri interprettions whih provided us deeper understnding of our lgebri findings. Copyright 008 Lrry Lrson & Slly Keely. Originlly written: 008-0-14 All Rights Reserved. Lst revision: 009-03-9