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 .



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 $0$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

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 $\;(-\infty,-4]\cup[0,+\infty)$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.