Question Number 53732 by ajfour last updated on 25/Jan/19
$${f}\left({x}\right)=\frac{\left({x}+{a}\right)\left({x}+{b}\right)}{\left({x}−{a}\right)\left({x}−{b}\right)} \\ $$$${Find}\:{minimum}\:{and}\:{maximum}. \\ $$
Commented by mr W last updated on 25/Jan/19
$${at}\:{x}={a}\:{and}\:{x}={b}:\:{f}\left({x}\right)\rightarrow\pm\infty \\ $$
Commented by ajfour last updated on 25/Jan/19
$${There}\:{is}\:{a}\:{definite}\:{answer}\:{given}, \\ $$$${maximum}\:{and}\:{minimum}\:{are} \\ $$$${asked}.\:{Its}\:{from}\:{a}\:{good}\:{book},\:{Sir}. \\ $$$${must}\:{then}\:{be}\:{local}\:{minimum}\:{and} \\ $$$${maximum}. \\ $$
Commented by mr W last updated on 25/Jan/19
$${since}\:{f}\left({x}\right)\rightarrow\pm\infty\:{when}\:{x}\rightarrow{a}\:{or}\:\rightarrow{b}, \\ $$$${there}\:{is}\:{no}\:{min}.\:{or}\:{max}.\:{but} \\ $$$${there}\:{can}\:{be}\:{a}\:{local}\:{minimum}\:{and}\:{a} \\ $$$${local}\:{maximum}. \\ $$$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} +\left({a}+{b}\right){x}+{ab}}{{x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}} \\ $$$${f}\left({x}\right)=\mathrm{1}+\frac{\mathrm{2}\left({a}+{b}\right){x}}{{x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}} \\ $$$${f}'\left({x}\right)=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}}−\frac{{x}\left(\mathrm{2}{x}−\left({a}+{b}\right)\right)}{\left({x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}={x}\left(\mathrm{2}{x}−\left({a}+{b}\right)\right) \\ $$$${x}^{\mathrm{2}} ={ab} \\ $$$$\Rightarrow{x}=\pm\sqrt{{ab}} \\ $$$$ \\ $$$${i}.{e}.\:{if}\:{ab}\geqslant\mathrm{0}\:{there}\:{are}\:{local}\:{min}.\:{and} \\ $$$${local}\:{max}.\:{at}\:{x}=\pm\sqrt{{ab}} \\ $$$${f}\left({x}\right)_{{min}/{max}} =\frac{\left({a}\pm\sqrt{{ab}}\right)\left({b}\pm\sqrt{{ab}}\right)}{\left({a}\mp\sqrt{{ab}}\right)\left({b}\mp\sqrt{{ab}}\right)} \\ $$$$ \\ $$$${if}\:{ab}<\mathrm{0}\:{there}\:{is}\:{no}\:{local}\:{min}.\:{or}\:{max}. \\ $$$$ \\ $$$${but}\:{we}\:{don}'{t}\:{need}\:{to}\:{use}\:{calculus}\:{to} \\ $$$${get}\:{this}\:{result},\:{just}\:{look}\:{at} \\ $$$${f}\left({x}\right)=\mathrm{1}+\frac{\mathrm{2}\left({a}+{b}\right){x}}{{x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}}=\mathrm{1}+\frac{\mathrm{2}\left({a}+{b}\right)}{{x}+\frac{{ab}}{{x}}−\left({a}+{b}\right)} \\ $$$${if}\:{ab}\geqslant\mathrm{0}, \\ $$$${x}+\frac{{ab}}{{x}}\geqslant\mathrm{2}\sqrt{{ab}} \\ $$$${x}+\frac{{ab}}{{x}}\leqslant−\mathrm{2}\sqrt{{ab}} \\ $$$${f}\left({x}\right)=\mathrm{1}+\frac{\mathrm{2}\left({a}+{b}\right)}{{x}+\frac{{ab}}{{x}}−\left({a}+{b}\right)}\leqslant\mathrm{1}+\frac{\mathrm{2}\left({a}+{b}\right)}{\mathrm{2}\sqrt{{ab}}−\left({a}+{b}\right)}=−\left(\frac{\sqrt{{a}}+\sqrt{{b}}}{\sqrt{{a}}−\sqrt{{b}}}\right)^{\mathrm{2}} ={f}_{{max}} \\ $$$${f}\left({x}\right)=\mathrm{1}+\frac{\mathrm{2}\left({a}+{b}\right)}{{x}+\frac{{ab}}{{x}}−\left({a}+{b}\right)}\geqslant\mathrm{1}+\frac{\mathrm{2}\left({a}+{b}\right)}{−\mathrm{2}\sqrt{{ab}}−\left({a}+{b}\right)}=−\left(\frac{\sqrt{{a}}−\sqrt{{b}}}{\sqrt{{a}}+\sqrt{{b}}}\right)^{\mathrm{2}} ={f}_{{min}} \\ $$
Commented by ajfour last updated on 25/Jan/19
$${answer}:\:\:−\left(\frac{\sqrt{{a}}−\sqrt{{b}}}{\sqrt{{a}}+\sqrt{{b}}}\right)^{\mathrm{2}} \:\&\:\:−\left(\frac{\sqrt{{a}}+\sqrt{{b}}}{\sqrt{{a}}−\sqrt{{b}}}\right)^{\mathrm{2}} \:. \\ $$
Commented by mr W last updated on 25/Jan/19
$${this}\:{is}\:{the}\:{same}\:{as}\:{my}\:{result}\:\left({see}\:{below}\right). \\ $$$${answer}\:{in}\:{book}\:{is}\:{not}\:{the}\:{best},\:{since} \\ $$$${it}\:{requests}\:{that}\:{a}\geqslant\mathrm{0}\:{and}\:{b}\geqslant\mathrm{0},\:{but}\:{in} \\ $$$${fact}\:{this}\:{is}\:{not}\:{necessary}.\:{it}\:{is}\:{only} \\ $$$${necessary}\:{that}\:{ab}\geqslant\mathrm{0},\:{e}.{g}.\:{a}=−\mathrm{2},\:{b}=−\mathrm{5}. \\ $$
Commented by mr W last updated on 25/Jan/19
$$\frac{\left({a}+\sqrt{{ab}}\right)\left({b}+\sqrt{{ab}}\right)}{\left({a}−\sqrt{{ab}}\right)\left({b}−\sqrt{{ab}}\right)} \\ $$$$=\frac{\sqrt{{a}}\left(\sqrt{{a}}+\sqrt{{b}}\right)\sqrt{{b}}\left(\sqrt{{b}}+\sqrt{{a}}\right)}{\sqrt{{a}}\left(\sqrt{{a}}−\sqrt{{b}}\right)\sqrt{{b}}\left(\sqrt{{b}}−\sqrt{{a}}\right)} \\ $$$$=−\frac{\left(\sqrt{{a}}+\sqrt{{b}}\right)\left(\sqrt{{a}}+\sqrt{{b}}\right)}{\left(\sqrt{{a}}−\sqrt{{b}}\right)\left(\sqrt{{a}}−\sqrt{{b}}\right)} \\ $$$$=−\left(\frac{\sqrt{{a}}+\sqrt{{b}}}{\sqrt{{a}}−\sqrt{{b}}}\right)^{\mathrm{2}} ={as}\:{given}\:{in}\:{book} \\ $$$$ \\ $$$$\frac{\left({a}−\sqrt{{ab}}\right)\left({b}−\sqrt{{ab}}\right)}{\left({a}+\sqrt{{ab}}\right)\left({b}+\sqrt{{ab}}\right)} \\ $$$$=\frac{\sqrt{{a}}\left(\sqrt{{a}}−\sqrt{{b}}\right)\sqrt{{b}}\left(\sqrt{{b}}−\sqrt{{a}}\right)}{\sqrt{{a}}\left(\sqrt{{a}}+\sqrt{{b}}\right)\sqrt{{b}}\left(\sqrt{{b}}+\sqrt{{a}}\right)} \\ $$$$=−\frac{\left(\sqrt{{a}}−\sqrt{{b}}\right)\left(\sqrt{{a}}−\sqrt{{b}}\right)}{\left(\sqrt{{a}}+\sqrt{{b}}\right)\left(\sqrt{{a}}+\sqrt{{b}}\right)} \\ $$$$=−\left(\frac{\sqrt{{a}}−\sqrt{{b}}}{\sqrt{{a}}+\sqrt{{b}}}\right)^{\mathrm{2}} ={as}\:{given}\:{in}\:{book} \\ $$
Commented by ajfour last updated on 25/Jan/19
$${yes},\:{sir}.\:{too}\:{well}\:{did}\:{you}\:{explain}. \\ $$$${Thank}\:{you}\:{too}\:{much}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19
![y=(((x+a)(x+b))/((x−a)(x−b))) lny=ln(x+a)+ln(x+b)−ln(x−a)−ln(x−b) (1/y)(dy/dx)=(1/(x+a))+(1/(x+b))−(1/(x−a))−(1/(x−b)) for max/min (dy/dx)=0 (1/(x+a))−(1/(x−b))+(1/(x+b))−(1/(x−a))=0 ((x−b−x−a)/((x+a)(x−b)))+((x−a−x−b)/((x+b)(x−a)))=0 ((−(a+b))/((x+a)(x−b)))+((−(a+b))/((x+b)(x−a)))=0 (1/((x+a)(x−b)))+(1/((x+b)(x−a)))=0 x^2 −ax+bx−ab+x^2 −bx+ax−ab=0 2x^2 −2ab=0 x=±(√(ab)) (1/y)(dy/dx)=(1/(x+a))+(1/(x+b))−(1/(x−a))−(1/(x−b)) (dy/dx)=(((x+a)(x+b))/((x−a)(x−b)))[(1/(x+a))+(1/(x+b))−(1/(x−a))−(1/(x−b))] ((wait...)/)wl now i am going to find change of sign using first derivativd method to find max/min ((dy/dx))=(((x+a)(x+b))/(x^2 −x(a+b)+ab))×[(1/(x+a))−(1/(x−b))+(1/(x+b))−(1/(x−a))] =((x^2 +x(a+b)+ab)/(x^2 −x(a+b)+ab))×[((x−b−x−a)/(x^2 −xb+ax−ab))+((x−a−x−b)/(x^2 −ax+bx−ab))] =(N/D)×[((−(a+b))/(x^2 −ab+x(a−b)))+((−(a+b))/(x^2 −ab−x(a−b)))] =(N/D)×−(a+b)×[((x^2 −ab−x(a−b)+x^2 −ab+x(a−b))/((x^2 −ab)^2 −x^2 (a−b)^2 ))] nowD<N so(N/D)>1 (N/D)×−(a+b)=−ve [((2(x^2 −ab))/((x^2 −ab)^2 −x^2 {(a+b)^2 −4ab))] when x>(√(ab)) x^2 =h+ab =(N/D)×−(a+b)×[((2(x^2 −ab))/((x^2 −ab)^2 −x^2 {(a+b)^2 −4ab))] =(−ve)×[((2h)/(h^2 −(h+ab)(a−b)^2 ))]=(−ve)×((+ve)/(−ve)) when x>(√(ab)) sign change from −ve to +ve so at x=(√(ab)) f(x) min y_ =(((x+a)(x+b))/((x−a)(x−b))) (at x=(√(ab)) ) =(((√a) ((√b) +(√a) )×(√b) ((√a) +(√b) ))/((√a) ((√b) −(√a) )×(√b) ((√a) −(√b) ))) =(−)((((√a) +(√b) )^2 )/(((√(b−))(√a) )^2 )) →min value so at x=(√(ab)) min value at x=−(√(ab)) max value y=(((−(√(ab)) +a)(−(√(ab)) +b))/((−(√(ab)) −a)(−(√(ab)) −b))) =(((√a) ((√a) −(√b) )×−(√b) ((√a) −(√b) ))/(−(√a) ((√a) +(√b) )×−(√b) ((√a) +(√b) ))) =(−1)×((((√a) −(√b) )^2 )/(((√a) +(√b) )))←max vzlud...](Q53742.png)
$${y}=\frac{\left({x}+{a}\right)\left({x}+{b}\right)}{\left({x}−{a}\right)\left({x}−{b}\right)} \\ $$$${lny}={ln}\left({x}+{a}\right)+{ln}\left({x}+{b}\right)−{ln}\left({x}−{a}\right)−{ln}\left({x}−{b}\right) \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{x}+{a}}+\frac{\mathrm{1}}{{x}+{b}}−\frac{\mathrm{1}}{{x}−{a}}−\frac{\mathrm{1}}{{x}−{b}} \\ $$$${for}\:{max}/{min}\:\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{x}+{a}}−\frac{\mathrm{1}}{{x}−{b}}+\frac{\mathrm{1}}{{x}+{b}}−\frac{\mathrm{1}}{{x}−{a}}=\mathrm{0} \\ $$$$\frac{{x}−{b}−{x}−{a}}{\left({x}+{a}\right)\left({x}−{b}\right)}+\frac{{x}−{a}−{x}−{b}}{\left({x}+{b}\right)\left({x}−{a}\right)}=\mathrm{0} \\ $$$$\frac{−\left({a}+{b}\right)}{\left({x}+{a}\right)\left({x}−{b}\right)}+\frac{−\left({a}+{b}\right)}{\left({x}+{b}\right)\left({x}−{a}\right)}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\left({x}+{a}\right)\left({x}−{b}\right)}+\frac{\mathrm{1}}{\left({x}+{b}\right)\left({x}−{a}\right)}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{ax}+{bx}−{ab}+{x}^{\mathrm{2}} −{bx}+{ax}−{ab}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{ab}=\mathrm{0} \\ $$$${x}=\pm\sqrt{{ab}}\: \\ $$$$\frac{\mathrm{1}}{{y}}\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{x}+{a}}+\frac{\mathrm{1}}{{x}+{b}}−\frac{\mathrm{1}}{{x}−{a}}−\frac{\mathrm{1}}{{x}−{b}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\left({x}+{a}\right)\left({x}+{b}\right)}{\left({x}−{a}\right)\left({x}−{b}\right)}\left[\frac{\mathrm{1}}{{x}+{a}}+\frac{\mathrm{1}}{{x}+{b}}−\frac{\mathrm{1}}{{x}−{a}}−\frac{\mathrm{1}}{{x}−{b}}\right] \\ $$$$\frac{{wait}...}{}{wl} \\ $$$${now}\:{i}\:{am}\:{going}\:{to}\:{find}\:{change}\:{of}\:{sign} \\ $$$${using}\:{first}\:{derivativd}\:{method}\:{to}\:{find}\:{max}/{min} \\ $$$$\left(\frac{{dy}}{{dx}}\right)=\frac{\left({x}+{a}\right)\left({x}+{b}\right)}{{x}^{\mathrm{2}} −{x}\left({a}+{b}\right)+{ab}}×\left[\frac{\mathrm{1}}{{x}+{a}}−\frac{\mathrm{1}}{{x}−{b}}+\frac{\mathrm{1}}{{x}+{b}}−\frac{\mathrm{1}}{{x}−{a}}\right] \\ $$$$=\frac{{x}^{\mathrm{2}} +{x}\left({a}+{b}\right)+{ab}}{{x}^{\mathrm{2}} −{x}\left({a}+{b}\right)+{ab}}×\left[\frac{{x}−{b}−{x}−{a}}{{x}^{\mathrm{2}} −{xb}+{ax}−{ab}}+\frac{{x}−{a}−{x}−{b}}{{x}^{\mathrm{2}} −{ax}+{bx}−{ab}}\right] \\ $$$$=\frac{{N}}{{D}}×\left[\frac{−\left({a}+{b}\right)}{{x}^{\mathrm{2}} −{ab}+{x}\left({a}−{b}\right)}+\frac{−\left({a}+{b}\right)}{{x}^{\mathrm{2}} −{ab}−{x}\left({a}−{b}\right)}\right] \\ $$$$=\frac{{N}}{{D}}×−\left({a}+{b}\right)×\left[\frac{{x}^{\mathrm{2}} −{ab}−{x}\left({a}−{b}\right)+{x}^{\mathrm{2}} −{ab}+{x}\left({a}−{b}\right)}{\left({x}^{\mathrm{2}} −{ab}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \left({a}−{b}\right)^{\mathrm{2}} }\right] \\ $$$${nowD}<{N}\:\:{so}\frac{{N}}{{D}}>\mathrm{1}\:\:\:\frac{{N}}{{D}}×−\left({a}+{b}\right)=−{ve} \\ $$$$\left[\frac{\mathrm{2}\left({x}^{\mathrm{2}} −{ab}\right)}{\left({x}^{\mathrm{2}} −{ab}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \left\{\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{4}{ab}\right.}\right] \\ $$$${when}\:{x}>\sqrt{{ab}}\:\:{x}^{\mathrm{2}} ={h}+{ab} \\ $$$$=\frac{{N}}{{D}}×−\left({a}+{b}\right)×\left[\frac{\mathrm{2}\left({x}^{\mathrm{2}} −{ab}\right)}{\left({x}^{\mathrm{2}} −{ab}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \left\{\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{4}{ab}\right.}\right] \\ $$$$=\left(−{ve}\right)×\left[\frac{\mathrm{2}{h}}{{h}^{\mathrm{2}} −\left({h}+{ab}\right)\left({a}−{b}\right)^{\mathrm{2}} }\right]=\left(−{ve}\right)×\frac{+{ve}}{−{ve}} \\ $$$${when}\:{x}>\sqrt{{ab}}\:{sign}\:{change}\:{from}\:−{ve}\:{to}\:+{ve} \\ $$$${so}\:{at}\:{x}=\sqrt{{ab}}\:{f}\left({x}\right)\:{min} \\ $$$${y}_{} =\frac{\left({x}+{a}\right)\left({x}+{b}\right)}{\left({x}−{a}\right)\left({x}−{b}\right)}\:\left({at}\:{x}=\sqrt{{ab}}\:\right) \\ $$$$\:\:=\frac{\sqrt{{a}}\:\left(\sqrt{{b}}\:+\sqrt{{a}}\:\right)×\sqrt{{b}}\:\left(\sqrt{{a}}\:+\sqrt{{b}}\:\right)}{\sqrt{{a}}\:\left(\sqrt{{b}}\:−\sqrt{{a}}\:\right)×\sqrt{{b}}\:\left(\sqrt{{a}}\:−\sqrt{{b}}\:\right)} \\ $$$$=\left(−\right)\frac{\left(\sqrt{{a}}\:+\sqrt{{b}}\:\right)^{\mathrm{2}} }{\left(\sqrt{{b}−}\sqrt{{a}}\:\right)^{\mathrm{2}} }\:\rightarrow{min}\:{value} \\ $$$${so}\:{at}\:{x}=\sqrt{{ab}}\:\:{min}\:{value} \\ $$$$ \\ $$$${at}\:{x}=−\sqrt{{ab}}\:{max}\:{value} \\ $$$${y}=\frac{\left(−\sqrt{{ab}}\:+{a}\right)\left(−\sqrt{{ab}}\:+{b}\right)}{\left(−\sqrt{{ab}}\:−{a}\right)\left(−\sqrt{{ab}}\:−{b}\right)} \\ $$$$=\frac{\sqrt{{a}}\:\left(\sqrt{{a}}\:−\sqrt{{b}}\:\right)×−\sqrt{{b}}\:\left(\sqrt{{a}}\:−\sqrt{{b}}\:\right)}{−\sqrt{{a}}\:\left(\sqrt{{a}}\:+\sqrt{{b}}\:\right)×−\sqrt{{b}}\:\left(\sqrt{{a}}\:+\sqrt{{b}}\:\right)}\: \\ $$$$=\left(−\mathrm{1}\right)×\frac{\left(\sqrt{{a}}\:−\sqrt{{b}}\:\right)^{\mathrm{2}} }{\left(\sqrt{{a}}\:+\sqrt{{b}}\:\right)}\leftarrow{max}\:{vzlud}... \\ $$$$ \\ $$
Commented by ajfour last updated on 25/Jan/19
$${welcome}\:{back}\:{Sir}.. \\ $$
Commented by ajfour last updated on 25/Jan/19
$${thanks}\:{sir},\:{do}\:{i}\:{post}\:{the}\:{answer}? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19
$${i}\:{went}\:{to}\:{see}\:{the}\:{news}... \\ $$