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May 5th, 2016, 05:21 AM  #1 
Newbie Joined: May 2016 From: italy Posts: 22 Thanks: 0  how to find global maximum on open interval
hello, guys, I have a real problem to understand how to find the global maximum of this function (n+5)/radical (n^2+n) on an open interval I studied a lot about it but I could not find something useful about it. I know I should use limit as x approaches to +or infinity but I confused it with the horizontal asymptote if the limit exists that mean there is a global maximum or minimum I would be appreciated if someone explain me clearly I really need this one for solve supremum problem thank you. 
May 5th, 2016, 08:42 AM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677  Quote:
what "open interval" do you mean? why is the function in terms of $n$, then you mention limits as $x \to \pm \infty$ ? have you determined the domain of this function? what happens to the function values as $x \to 0^+$ and as $x \to 1^$ ?  
May 5th, 2016, 09:42 AM  #3  
Newbie Joined: May 2016 From: italy Posts: 22 Thanks: 0  Quote:
I wanted to see what will happen to my function as I will approach $x \to \pm \infty$ the domain of my function should be n(n+1)>=0 but how to find the horizontal asymptote I mean if I can find horizontal and vertical asymptote at least I can understand the behavior of my function and find x and yintercept so if I take the derivative of my function and set it equal to zero and find that possible point of changing the sign I can find which is maximum and which is minimum point but which one of them is global maximum and how can I know my function allowed me to find the global maximum or no  
May 5th, 2016, 10:14 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677 
$f(x)=\dfrac{x+5}{\sqrt{x^2+x}}$ when you analyze a function, the best place to start is the domain ... $x(x+1) > 0 \implies$ function domain is $(\infty,1) \cup (0,\infty)$ so, no yintecept, xintercept at $x=5$ horizontal asymptotes ... first note the fact that $\sqrt{x^2} = x$ ... $\displaystyle \lim_{x \to \infty} \dfrac{x+5}{\sqrt{x^2+x}}$ $\displaystyle \lim_{x \to \infty} \dfrac{\frac{x}{x}+\frac{5}{x}}{\sqrt{\frac{x^ 2}{x^2}+\frac{x}{x^2}}}$ $\displaystyle \lim_{x \to \infty} \dfrac{\frac{x}{x}+\frac{5}{x}}{\sqrt{1+\frac{ x}{x^2}}} = \dfrac{1+0}{\sqrt{1+0}} = 1$ horizontal asymptote at $y=1$ I'll let you determine the limit as $x \to +\infty$ to get the other horizontal asymptote. $f(x)$ is undefined at $x=0$ and$x=1$, so vertical asymptotes at $x=0$ and $x=1$ ... $\displaystyle \lim_{x \to 0^+} \dfrac{x+5}{\sqrt{x^2+x}} = \infty$ $\displaystyle \lim_{x \to 1^} \dfrac{x+5}{\sqrt{x^2+x}} = \infty$ no global maximum $f'(x) = \dfrac{4x}{(x^2+x)^{3/2}}$ no critical values in the domain ... $x < 1 \implies f'(x) > 0$ function is strictly increasing $x > 0 \implies f'(x) < 0$ function is strictly decreasing 
May 5th, 2016, 10:29 AM  #5  
Newbie Joined: May 2016 From: italy Posts: 22 Thanks: 0  Quote:
but why you let limit to approach 0 and 1 from left and write because you want to see what will happen at the end point or  
May 5th, 2016, 10:42 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677  when the function approaches a vertical asymptote, it has one of two ways to go ... up to $+\infty$ or down to $\infty$ ... nice to know if you have to sketch the graph.

May 5th, 2016, 11:04 AM  #7 
Newbie Joined: May 2016 From: italy Posts: 22 Thanks: 0  

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