How-may-we-plot-the-graph-of-f-x-x-x-x-2-x-1-with-the-help-of-a-variation-table-

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 + \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