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


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

$How may we plot the graph of$ $f (x) = x + \sqrt{\frac{x (x + 2)}{x + 1}}, with$ $the help of a variation table ?$
Commented by Ar Brandon last updated on 19/Jul/20
$a\ f(x)=x+(√((x(x+2))/(x+1))) i\Df; ((x(x+2))/((x+1)))≥0 , x≠−1 ⇒((x(x+1)(x+2))/((x+1)^2 ))≥0 ⇒x∈[−2,−1[∪[0,+∞[ ii\ lim_(x→−2^+ ) f(x)=−2 ,lim_(x→−1^− ) f(x)=+∞ ,lim_(x→0^+ ) f(x)=0 ,lim_(x→+∞) f(x)=+∞ iii\Asymptotes; x+1≠0 ⇒ x=−1 is a vertical asymptote. f(x)=x+((g(x))/(h(x))) ⇒ y=x is an oblique asymptote. iv\ f(x)=0⇒x+(√((x(x+2))/(x+1)))=0 ⇒((x(x+2))/(x+1))=x^2 ⇒x(x+2)=x^3 +x^2 ⇒x(x^2 −2)=0 ⇒x=−(√2) , x=0 , x=(√2) , x=(√2) ⇏f(x)=0 ⇒x=−(√2), x=0 ⇒(−(√2),0) (0,0) v\ f ′(x)=(x+(√(x+1−(1/(x+1)))))^′ =1+(1/2)∙(1/(√(x+1−(1/(x+1)))))∙(1+(1/((x+1)^2 ))) =1+(1/2)∙(1/(√((x^2 +2x)/(x+1))))∙(((x^2 +2x+2)/((x+1)^2 ))) f ′(x)=0 ⇒(√((x+1)/(x^2 +2x)))∙((x^2 +2x+2)/((x+1)^2 ))=−2 ⇒ (((x^2 +2x+2)^2 )/(x(x+2)(x+1)^3 ))=4 determinant ((x,(−2 −(√2)),( −1),( 0),( +∞)),((f ′(x)),( +),( +),( −),( −)),((f(x)),( _(−2) ↗^0 ),( _0 ↗^(+∞) ),( ^(+∞) ↘_0 ),( ^( 0) ↘_(−∞) )))$
$a ∖ f (x) = x + \sqrt{\frac{x (x + 2)}{x + 1}}$ $i ∖ D f; \frac{x (x + 2)}{(x + 1)} ⩾ 0, x \neq - 1 \Rightarrow \frac{x (x + 1) (x + 2)}{{(x + 1)}^{2}} ⩾ 0$ $\Rightarrow x \in [- 2, - 1 [\cup [0, + \infty [$ $i i ∖ \underset{x \to - 2^{+}}{\lim f} (x) = - 2, \underset{x \to - 1^{-}}{\lim f} (x) = + \infty, \underset{x \to 0^{+}}{\lim f} (x) = 0, \underset{x \to + \infty}{\lim f} (x) = + \infty$ $i i i ∖ Asymptotes; x + 1 \neq 0 \Rightarrow x = - 1 is a vertical asymptote .$ $f (x) = x + \frac{g (x)}{h (x)} \Rightarrow y = x is an oblique asymptote .$ $i v ∖ f (x) = 0 \Rightarrow x + \sqrt{\frac{x (x + 2)}{x + 1}} = 0 \Rightarrow \frac{x (x + 2)}{x + 1} = x^{2}$ $\Rightarrow x (x + 2) = x^{3} + x^{2} \Rightarrow x (x^{2} - 2) = 0$ $\Rightarrow x = - \sqrt{2}, x = 0, x = \sqrt{2}, x = \sqrt{2} ⇏ f (x) = 0$ $\Rightarrow x = - \sqrt{2}, x = 0 \Rightarrow (- \sqrt{2}, 0) (0, 0)$ $v ∖ f^{'} (x) = {(x + \sqrt{x + 1 - \frac{1}{x + 1}})}^{^{'}} = 1 + \frac{1}{2} \cdot \frac{1}{\sqrt{x + 1 - \frac{1}{x + 1}}} \cdot (1 + \frac{1}{{(x + 1)}^{2}})$ $= 1 + \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{x^{2} + 2 x}{x + 1}}} \cdot (\frac{x^{2} + 2 x + 2}{{(x + 1)}^{2}})$ $f^{'} (x) = 0 \Rightarrow \sqrt{\frac{x + 1}{x^{2} + 2 x}} \cdot \frac{x^{2} + 2 x + 2}{{(x + 1)}^{2}} = - 2 \Rightarrow \frac{{(x^{2} + 2 x + 2)}^{2}}{x (x + 2) {(x + 1)}^{3}} = 4$ $\| \begin{matrix} x & - 2 - \sqrt{2} & - 1 & 0 & + \infty \\ f^{'} (x) & + & + & - & - \\ f (x) & \underset{- 2}{} ↗^{0} & \underset{0}{} ↗^{+ \infty} & \overset{+ \infty}{} ↘_{0} & \overset{0}{} ↘_{- \infty} \end{matrix} \|$
Commented by Ar 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 by Ar Brandon last updated on 19/Jul/20

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