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 4820 by love math last updated on 16/Mar/16

y=f(x)+g(x) f(x) − odd function g(x) − even function find f(0), if y= 2x^2 +((sin x)/3)+1

$${y}={f}\left({x}\right)+{g}\left({x}\right) \\ $$$$ \\ $$$${f}\left({x}\right)\:−\:{odd}\:{function} \\ $$$${g}\left({x}\right)\:−\:{even}\:{function} \\ $$$$ \\ $$$${find}\:{f}\left(\mathrm{0}\right),\:{if}\:{y}=\:\mathrm{2}{x}^{\mathrm{2}} +\frac{{sin}\:{x}}{\mathrm{3}}+\mathrm{1} \\ $$

Answered by prakash jain last updated on 16/Mar/16

y(x)+y(−x)=f(x)+g(x)+f(−x)+g(−x) =f(x)+g(x)+f(x)−g(x)=2f(x) y(x)+y(−x)=2x^2 +((sin x)/3)+1+2(−x)^2 +((sin (−x))/3)+1 =4x^2 +2=2f(x) f(x)=2x^2 +1 f(0)=1

$${y}\left({x}\right)+{y}\left(−{x}\right)={f}\left({x}\right)+{g}\left({x}\right)+{f}\left(−{x}\right)+{g}\left(−{x}\right) \\ $$$$={f}\left({x}\right)+{g}\left({x}\right)+{f}\left({x}\right)−{g}\left({x}\right)=\mathrm{2}{f}\left({x}\right) \\ $$$${y}\left({x}\right)+{y}\left(−{x}\right)=\mathrm{2}{x}^{\mathrm{2}} +\frac{\mathrm{sin}\:{x}}{\mathrm{3}}+\mathrm{1}+\mathrm{2}\left(−{x}\right)^{\mathrm{2}} +\frac{\mathrm{sin}\:\left(−{x}\right)}{\mathrm{3}}+\mathrm{1} \\ $$$$=\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}=\mathrm{2}{f}\left({x}\right) \\ $$$${f}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$

Commented by Yozzii last updated on 16/Mar/16

f(x) is an odd function; 2x^2 +1 is even.

$${f}\left({x}\right)\:{is}\:{an}\:{odd}\:{function};\:\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\:{is}\:{even}. \\ $$

Commented by Rasheed Soomro last updated on 16/Mar/16

f(x) is an odd function, sof(−x) =−f(x) ∴ y(x)+y(−x)=2g(x)

$${f}\left({x}\right)\:{is}\:{an}\:{odd}\:{function},\:{sof}\left(−{x}\right)\:=−{f}\left({x}\right) \\ $$$$\therefore\:{y}\left({x}\right)+{y}\left(−{x}\right)=\mathrm{2}{g}\left({x}\right) \\ $$

Commented by prakash jain last updated on 16/Mar/16

Sorry. I switch f(x) and g(x). Took f(x) as even and g(x) as odd. So g(x)=2x^2 +1 f(x)=y(x)−g(x)=((sin x)/3) f(0)=0

$$\mathrm{Sorry}.\:\mathrm{I}\:\mathrm{switch}\:{f}\left({x}\right)\:\mathrm{and}\:{g}\left({x}\right). \\ $$$$\mathrm{Took}\:{f}\left({x}\right)\:{as}\:{even}\:{and}\:{g}\left({x}\right)\:{as}\:{odd}. \\ $$$$\mathrm{So}\: \\ $$$${g}\left({x}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}\left({x}\right)={y}\left({x}\right)−{g}\left({x}\right)=\frac{\mathrm{sin}\:{x}}{\mathrm{3}} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com