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Recently I got stuck on a calculus problem, where I have the following function :

$$f(x) = x + 1 + \sqrt{x^2 + 4x}$$

defined on $]-\infty,-4] \cup [0, +\infty[$

I need to find the sign of it's derivative (I mean when it's positive or negative) :

So I computed the drivative of $f$ and got this :

$$f'(x) = 1 + \frac{x+2}{\sqrt{x^2 + 4x}} $$

But I'm stuck at solving the inequality $f'(x) > 0$ or the equation $f'(x) = 0$ to determine it's sign (when it's positive/negative). How can I do that? Also more generally how can I solve most of those kind of inequalities? Is there a specific approach for that?

Also I would like to get something clarified that is confusing me for a while; The problem ask us to prove that $f(x)$ is derivable at $0$ and $-4$, I computed the $\lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}$ and I got $\infty$ for both $-4$ and $0$.

My question is that at first I computed the limit for $x \to ^> 0 $ and I got infinity is that enough to prove that $f$ is not derivable at $0$ or should I compute at $x \to ^< 0 $ too?

Thank's for your time .

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Hint

if $x>0\;\; f'(x)>0$

and for $x<-4$,

use $x^2+4x=(x+2)^2-4$ to get

$\sqrt{x^2+4x}<|x+2|$ and $f'(x)<0$.

for the question about derivative.

as the limits are infinite, the function is not differentiable neither at $-4$ nor at $0$.

you cannot compute the limit at $0^-$ since the function is not defined on the left side of $0$.

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  • $\begingroup$ Oh thank you so much, this make everything more clear, also what about my second problem, could you help me get rid of my confusion ? $\endgroup$ – Anis Souames Dec 5 '16 at 21:15
  • $\begingroup$ Thank's for your explanation on the 2nd problem, but if it was defined on both side of 0 can I do what I said in the question ? Computing only on one side and see ? $\endgroup$ – Anis Souames Dec 5 '16 at 21:25
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    $\begingroup$ @AnisSouames if you find infinity as the limit at one side, you can say it is not differentiable. $\endgroup$ – hamam_Abdallah Dec 5 '16 at 21:30
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First, find the domain of the inequation: it is defined by the condition $x^2+4x=x(x+4)\ge 0\iff x\le -4\enspace\text{ or }\enspace x\ge0$.

  • If $x\ge 0$, it is clear that $\;\sqrt{x^2+4x}+x+1\ge 0$.
  • If $x\le -4$, the inequation is equivalent to $$\sqrt{x^2+4x}\ge -(x+1)\iff x^2+4\ge x^2+2x+1\iff x\le\frac32,$$ which is true.

Hence the solutions are the the domain of the inequation, i.e. the set $\;(-\infty,-4]\cup[0,+\infty)$.

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  • $\begingroup$ Thank you so much for this calrification, could you also clear up my confusion on the 2nd problem please, ? Thank's ! $\endgroup$ – Anis Souames Dec 5 '16 at 21:23
  • $\begingroup$ You mean derivability at $0$ and $4$? $\endgroup$ – Bernard Dec 5 '16 at 21:33

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