Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 204426 by universe last updated on 17/Feb/24

let a , b >0 find all differentiable function f:(0,∞)→(0,∞) such that f′((a/x)) = ((bx)/(f(x))) , ∀ x>0

$$\:\:\mathrm{let}\:\mathrm{a}\:,\:\mathrm{b}\:>\mathrm{0}\:\:\mathrm{find}\:\mathrm{all}\:\mathrm{differentiable}\:\mathrm{function} \\ $$$$\:\:\:\mathrm{f}:\left(\mathrm{0},\infty\right)\rightarrow\left(\mathrm{0},\infty\right)\:\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\:\:\:\mathrm{f}'\left(\frac{{a}}{\mathrm{x}}\right)\:\:=\:\:\frac{\mathrm{bx}}{\mathrm{f}\left(\mathrm{x}\right)}\:\:\:\:,\:\:\:\forall\:\mathrm{x}>\mathrm{0} \\ $$

Answered by MrGaster last updated on 03/Feb/25

Let f(x)=(√(bx^2 +c)),c=((a^2 b)/4) f′(x)=((bx)/( (√(bx^2 +c)))) Substitute x=(a/y)into f′(x): f′((a/y))=((b((a/y)))/( (√(b((a/y))^2 +c))))=(((ba)/y)/( (√(((ba^2 )/y^2 )+c)))) (((ba)/y)/( (√(((ba^2 )/y^2 )+c))))=((by)/(f(y))) Substitute f(y)=(√(by^2 +c)): (((by)/y)/( (√(((ba^2 )/y^2 )+c))))=((by)/(f(y))) Substitute f(y)=(√(by^2 +c)): ((ba)/(y/( (√(((ba^2 )/y^2 )+c)))))=((by)/( (√(by^2 +c)))) ((ba)/y)∙(√(by^2 +c))=by∙(√(((ba^2 )/y^2 )+c)) (((ba)/y))^2 (by^2 +c)=b^2 y^2 (((ba^2 )/y^2 )+c) ((b^2 a^2 )/y^2 )(by^2 +c)=b^2 y^2 (((ba^2 )/y^2 )+c) (a^2 /y^2 )(by^2 +c)=y^2 ((a^2 /y^2 )+c) a^2 b+((a^2 c)/y^2 )=a^2 +cy^2 Subtitute c=((a^2 b)/4): a^2 b+((a^2 ∙((a^2 b)/4))/y^2 )=a^2 +((a^2 b)/4)y^2 4a^2 by^2 +a^4 b=4a^2 y^2 +a^2 by^2 a^2 by^4 −4a^2 by^2 +a^4 b=0 y^4 −4y^2 +a^2 =0 Let z=y^2 : z^2 −4z+a^2 =0 z=2±(√(4−a^2 )) y=(√(2±(√(4−a^2 )))) so: f(x)=(√(bx^2 +((a^2 b)/4)))

$$\mathrm{Let}\:{f}\left({x}\right)=\sqrt{{bx}^{\mathrm{2}} +{c}},{c}=\frac{{a}^{\mathrm{2}} {b}}{\mathrm{4}} \\ $$$${f}'\left({x}\right)=\frac{{bx}}{\:\sqrt{{bx}^{\mathrm{2}} +{c}}} \\ $$$$\mathrm{Substitute}\:{x}=\frac{{a}}{{y}}\mathrm{into}\:{f}'\left({x}\right): \\ $$$${f}'\left(\frac{{a}}{{y}}\right)=\frac{{b}\left(\frac{{a}}{{y}}\right)}{\:\sqrt{{b}\left(\frac{{a}}{{y}}\right)^{\mathrm{2}} +{c}}}=\frac{\frac{{ba}}{{y}}}{\:\sqrt{\frac{{ba}^{\mathrm{2}} }{{y}^{\mathrm{2}} }+{c}}} \\ $$$$\frac{\frac{{ba}}{{y}}}{\:\sqrt{\frac{{ba}^{\mathrm{2}} }{{y}^{\mathrm{2}} }+{c}}}=\frac{{by}}{{f}\left({y}\right)} \\ $$$$\mathrm{Substitute}\:{f}\left({y}\right)=\sqrt{{by}^{\mathrm{2}} +{c}}: \\ $$$$\frac{\frac{{by}}{{y}}}{\:\sqrt{\frac{{ba}^{\mathrm{2}} }{{y}^{\mathrm{2}} }+{c}}}=\frac{{by}}{{f}\left({y}\right)} \\ $$$$\mathrm{Substitute}\:{f}\left({y}\right)=\sqrt{{by}^{\mathrm{2}} +{c}}: \\ $$$$\frac{{ba}}{\frac{{y}}{\:\sqrt{\frac{{ba}^{\mathrm{2}} }{{y}^{\mathrm{2}} }+{c}}}}=\frac{{by}}{\:\sqrt{{by}^{\mathrm{2}} +{c}}} \\ $$$$\frac{{ba}}{{y}}\centerdot\sqrt{{by}^{\mathrm{2}} +{c}}={by}\centerdot\sqrt{\frac{{ba}^{\mathrm{2}} }{{y}^{\mathrm{2}} }+{c}} \\ $$$$\left(\frac{{ba}}{{y}}\right)^{\mathrm{2}} \left({by}^{\mathrm{2}} +{c}\right)={b}^{\mathrm{2}} {y}^{\mathrm{2}} \left(\frac{{ba}^{\mathrm{2}} }{{y}^{\mathrm{2}} }+{c}\right) \\ $$$$\frac{{b}^{\mathrm{2}} {a}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\left({by}^{\mathrm{2}} +{c}\right)={b}^{\mathrm{2}} {y}^{\mathrm{2}} \left(\frac{{ba}^{\mathrm{2}} }{{y}^{\mathrm{2}} }+{c}\right) \\ $$$$\frac{{a}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\left({by}^{\mathrm{2}} +{c}\right)={y}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} }{{y}^{\mathrm{2}} }+{c}\right) \\ $$$${a}^{\mathrm{2}} {b}+\frac{{a}^{\mathrm{2}} {c}}{{y}^{\mathrm{2}} }={a}^{\mathrm{2}} +{cy}^{\mathrm{2}} \\ $$$$\mathrm{Subtitute}\:{c}=\frac{{a}^{\mathrm{2}} {b}}{\mathrm{4}}: \\ $$$${a}^{\mathrm{2}} {b}+\frac{{a}^{\mathrm{2}} \centerdot\frac{{a}^{\mathrm{2}} {b}}{\mathrm{4}}}{{y}^{\mathrm{2}} }={a}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} {b}}{\mathrm{4}}{y}^{\mathrm{2}} \\ $$$$\mathrm{4}{a}^{\mathrm{2}} {by}^{\mathrm{2}} +{a}^{\mathrm{4}} {b}=\mathrm{4}{a}^{\mathrm{2}} {y}^{\mathrm{2}} +{a}^{\mathrm{2}} {by}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} {by}^{\mathrm{4}} −\mathrm{4}{a}^{\mathrm{2}} {by}^{\mathrm{2}} +{a}^{\mathrm{4}} {b}=\mathrm{0} \\ $$$${y}^{\mathrm{4}} −\mathrm{4}{y}^{\mathrm{2}} +{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{Let}\:{z}={y}^{\mathrm{2}} : \\ $$$${z}^{\mathrm{2}} −\mathrm{4}{z}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$${z}=\mathrm{2}\pm\sqrt{\mathrm{4}−{a}^{\mathrm{2}} } \\ $$$${y}=\sqrt{\mathrm{2}\pm\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }} \\ $$$$\mathrm{so}: \\ $$$${f}\left({x}\right)=\sqrt{{bx}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} {b}}{\mathrm{4}}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com