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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 Tags find, global, interval, maximum, open Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post One Calculus 5 April 10th, 2013 09:56 AM Singularity Real Analysis 1 February 15th, 2012 02:21 PM Pumpkin99 Algebra 7 November 16th, 2011 09:02 AM ArmiAldi Calculus 2 March 12th, 2008 08:03 PM cos5000 Calculus 4 December 9th, 2007 07:58 PM

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