October 5th, 2014, 07:56 PM  #1 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,428 Thanks: 118  Maximum and Minimum Value of a Fraction Function
Given a function $\displaystyle f(x)=\frac{2x^2}{x+1}$. Determine its minimum and maximum value. I found the first derivative as $\displaystyle f'(x)=\frac{2x^2+4x}{(x+1)^2}$, thus $\displaystyle 2x^2+4x=0$ and therefore I got $\displaystyle x_1=0$ and $\displaystyle x_2=2$. How to determine its minimum and maximum value without using second derivative? Should I just substitute both x values to f(x) then conclude the lesser one as minimum and the bigger one as maximum based on their results? 
October 5th, 2014, 08:26 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs 
A precalculus technique would be to write the equation as a quadratic in $\displaystyle x$: $\displaystyle 2x^2yxy=0$ Set the discriminant greater than or equal to 0: $\displaystyle (y)^24(2)(y)\ge0$ $\displaystyle y(y+8)\ge0$ The solution for $\displaystyle y$ is: $\displaystyle (\infty,8]\,\cup\,[0,\infty)$ Hence. the given rational function must have a local maximum of $\displaystyle y=8$ and a local minimum of $\displaystyle y=0$. Last edited by MarkFL; October 8th, 2014 at 12:06 AM. 
October 5th, 2014, 09:50 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,031 Thanks: 2342 Math Focus: Mainly analysis and algebra 
$$\lim_{x \to \infty} \frac{2x^2}{x+1} = \lim_{x \to \infty} \frac{2x}{1+\tfrac1x} = \infty \\ \lim_{x \to \infty} \frac{2x^2}{x+1} = \lim_{x \to \infty} \frac{2x}{1+\tfrac1x} = \infty$$ So the maximum and minimum values are $\pm \infty$, or rather, $f(x)$ is unbounded above and below. We can get the same result by considering the two onesided limits at $x=1$. Last edited by greg1313; October 6th, 2014 at 12:15 AM. 
October 6th, 2014, 07:58 AM  #4 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,428 Thanks: 118 
Whoa, I intended to use this question on a textbook I am working on, but since the solution is infinity, guess I have to delete or change it.

October 6th, 2014, 08:21 AM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,658 Thanks: 651 Math Focus: Wibbly wobbly timeywimey stuff.  
October 7th, 2014, 11:56 PM  #6 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,428 Thanks: 118  
October 8th, 2014, 12:07 AM  #7 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs  
October 8th, 2014, 04:13 AM  #8 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,428 Thanks: 118  
October 8th, 2014, 06:51 AM  #9  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs  Quote:
We find: $\displaystyle \lim_{x\to\infty}=\infty$ $\displaystyle \lim_{x\to1^{}}f(x)=\infty$ $\displaystyle \lim_{x\to1^{+}}f(x)=\infty$ $\displaystyle \lim_{x\to\infty}=\infty$  
October 8th, 2014, 04:35 PM  #10  
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,428 Thanks: 118  Quote:
 

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fraction, function, maximum, minimum 
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