Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 165552 by mnjuly1970 last updated on 03/Feb/22

f(x)= (( x +m)/(∣x∣ + 6)) and f is strictly monoton .find the value(s) of m . m∈Z

$$ \\ $$$$\:\:\:\:{f}\left({x}\right)=\:\frac{\:{x}\:+{m}}{\mid{x}\mid\:+\:\mathrm{6}}\:\:{and}\:\:{f}\:{is} \\ $$$$\:{strictly}\:\:{monoton}\:.{find}\:{the}\:{value}\left({s}\right) \\ $$$$\:\:{of}\:\:\:{m}\:\:\:.\:\:{m}\in\mathbb{Z} \\ $$

Answered by TheSupreme last updated on 03/Feb/22

f′(x)>0 ((∣x∣+m−sgn(x)(x+m))/((∣x∣+6)^2 ))=((∣x∣+m−∣x∣−sgn(x)m)/((∣x∣+6)^2 ))>0→m(1−sgn(x))>0 ∄m ∣ ∀x∈R: f(x) is strictly monoton f(x) is monoton for m=0 → f(x)=sgn(x)

$${f}'\left({x}\right)>\mathrm{0} \\ $$$$\frac{\mid{x}\mid+{m}−{sgn}\left({x}\right)\left({x}+{m}\right)}{\left(\mid{x}\mid+\mathrm{6}\right)^{\mathrm{2}} }=\frac{\mid{x}\mid+{m}−\mid{x}\mid−{sgn}\left({x}\right){m}}{\left(\mid{x}\mid+\mathrm{6}\right)^{\mathrm{2}} }>\mathrm{0}\rightarrow{m}\left(\mathrm{1}−{sgn}\left({x}\right)\right)>\mathrm{0} \\ $$$$\nexists{m}\:\mid\:\forall{x}\in\mathbb{R}:\:{f}\left({x}\right)\:{is}\:{strictly}\:{monoton} \\ $$$$ \\ $$$${f}\left({x}\right)\:{is}\:{monoton}\:{for}\:{m}=\mathrm{0}\:\rightarrow\:{f}\left({x}\right)={sgn}\left({x}\right) \\ $$

Answered by mahdipoor last updated on 03/Feb/22

⇒f= { ((((x+m)/(x+6)) x>0)),((((x+m)/(6−x)) x≤0)) :} ⇒f^′ = { ((((6−m)/((x+6)^2 )) x<0)),((((m+6)/((6−x)^2 )) x≥0)) :} { ((6−m<0 & m+6<0 ⇒ ∄)),((6−m>0 & m+6>0 ⇒ −6<m<6)) :} m∈Z⇒m∈{−5,−4,...,4,5}

$$\Rightarrow{f}=\begin{cases}{\frac{{x}+{m}}{{x}+\mathrm{6}}\:\:{x}>\mathrm{0}}\\{\frac{{x}+{m}}{\mathrm{6}−{x}}\:\:{x}\leqslant\mathrm{0}}\end{cases}\:\Rightarrow{f}^{'} =\begin{cases}{\frac{\mathrm{6}−{m}}{\left({x}+\mathrm{6}\right)^{\mathrm{2}} }\:\:{x}<\mathrm{0}}\\{\frac{{m}+\mathrm{6}}{\left(\mathrm{6}−{x}\right)^{\mathrm{2}} }\:\:{x}\geqslant\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{\mathrm{6}−{m}<\mathrm{0}\:\&\:{m}+\mathrm{6}<\mathrm{0}\:\Rightarrow\:\nexists}\\{\mathrm{6}−{m}>\mathrm{0}\:\&\:{m}+\mathrm{6}>\mathrm{0}\:\Rightarrow\:−\mathrm{6}<{m}<\mathrm{6}}\end{cases} \\ $$$${m}\in\mathrm{Z}\Rightarrow{m}\in\left\{−\mathrm{5},−\mathrm{4},...,\mathrm{4},\mathrm{5}\right\} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com