Question Number 104159 by Ar Brandon last updated on 19/Jul/20

How may we plot the graph of f(x)=x+(√((x(x+2))/(x+1))) , with the help of a variation table ?

Commented byAr Brandon last updated on 19/Jul/20

a\ f(x)=x+(√((x(x+2))/(x+1))) i\Df; ((x(x+2))/((x+1)))≥0 , x≠−1 ⇒((x(x+1)(x+2))/((x+1)^2 ))≥0 ⇒x∈[−2,−1[∪[0,+∞[ ii\ lim_(x→−2^+ ) f(x)=−2 ,lim_(x→−1^− ) f(x)=+∞ ,lim_(x→0^+ ) f(x)=0 ,lim_(x→+∞) f(x)=+∞ iii\Asymptotes; x+1≠0 ⇒ x=−1 is a vertical asymptote. f(x)=x+((g(x))/(h(x))) ⇒ y=x is an oblique asymptote. iv\ f(x)=0⇒x+(√((x(x+2))/(x+1)))=0 ⇒((x(x+2))/(x+1))=x^2 ⇒x(x+2)=x^3 +x^2 ⇒x(x^2 −2)=0 ⇒x=−(√2) , x=0 , x=(√2) , x=(√2) ⇏f(x)=0 ⇒x=−(√2), x=0 ⇒(−(√2),0) (0,0) v\ f ′(x)=(x+(√(x+1−(1/(x+1)))))^′ =1+(1/2)∙(1/(√(x+1−(1/(x+1)))))∙(1+(1/((x+1)^2 ))) =1+(1/2)∙(1/(√((x^2 +2x)/(x+1))))∙(((x^2 +2x+2)/((x+1)^2 ))) f ′(x)=0 ⇒(√((x+1)/(x^2 +2x)))∙((x^2 +2x+2)/((x+1)^2 ))=−2 ⇒ (((x^2 +2x+2)^2 )/(x(x+2)(x+1)^3 ))=4 determinant ((x,(−2 −(√2)),( −1),( 0),( +∞)),((f ′(x)),( +),( +),( −),( −)),((f(x)),( _(−2) ↗^0 ),( _0 ↗^(+∞) ),( ^(+∞) ↘_0 ),( ^( 0) ↘_(−∞) )))

Commented byAr Brandon last updated on 19/Jul/20

This is what I tried doing. But I′m not getting right. As this doesn′t represent f(x) Any idea, please ?

Commented byAr Brandon last updated on 19/Jul/20