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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
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 Originally Posted by programmer the global maximum of this function (n+5)/radical (n^2+n) on an open interval 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
where did you get this idea?

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
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 Originally Posted by skeeter where did you get this idea? 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 $x \to -1^-$ ?
I supposed n term and x term are equal they are just terms
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 y-intercept
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 y-intecept, x-intercept 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 Thanks from programmer
May 5th, 2016, 10:29 AM   #5
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 Originally Posted by skeeter $f(x)=\dfrac{x+5}{\sqrt{x^2+x}}$ 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 y-intecept, x-intercept 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
wow amazing answer thank you so much
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
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 Originally Posted by programmer wow amazing answer thank you so much 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
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
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 Originally Posted by skeeter 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.
really clear thank you so much you have to be one of that good college professor amazing man

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