My Math Forum Maximum and Minimum Value of a Fraction Function

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 October 5th, 2014, 06:56 PM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,818 Thanks: 123 Math Focus: Trigonometry and Logarithm 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, 07:26 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs A precalculus technique would be to write the equation as a quadratic in $\displaystyle x$: $\displaystyle 2x^2-yx-y=0$ Set the discriminant greater than or equal to 0: $\displaystyle (-y)^2-4(2)(-y)\ge0$ $\displaystyle y(y+8)\ge0$ 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 $\displaystyle y=0$. Thanks from topsquark Last edited by MarkFL; October 7th, 2014 at 11:06 PM.
 October 5th, 2014, 08:50 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,233 Thanks: 2411 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 one-sided limits at $x=-1$. Thanks from topsquark Last edited by greg1313; October 5th, 2014 at 11:15 PM.
 October 6th, 2014, 06:58 AM #4 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,818 Thanks: 123 Math Focus: Trigonometry and Logarithm 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, 07:21 AM   #5
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Quote:
 Originally Posted by Monox D. I-Fly 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.
You could set an interval. Or simply state that the problem is looking for relative maxima/minima.

-Dan

October 7th, 2014, 10:56 PM   #6
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Quote:
 Originally Posted by topsquark You could set an interval. Or simply state that the problem is looking for relative maxima/minima. -Dan
The relative maxima is 0 and the relative minima is -8, am I right?

October 7th, 2014, 11:07 PM   #7
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Quote:
 Originally Posted by Monox D. I-Fly The relative maxima is 0 and the relative minima is -8, am I right?
No, switch them.

October 8th, 2014, 03:13 AM   #8
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Quote:
 Originally Posted by MarkFL No, switch them.
H... How could the minima is bigger than the maxima? Minima is the lowest value and maxima is the biggest value, right?

October 8th, 2014, 05:51 AM   #9
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Quote:
 Originally Posted by Monox D. I-Fly H... How could the minima is bigger than the maxima? Minima is the lowest value and maxima is the biggest value, right?
The given function is split at $x=-1$ by a vertical asymptote. To the left is a piece that has a local maximum of -8 and to the right is a piece that has a local minimum of 0.

We find:

$\displaystyle \lim_{x\to-\infty}=-\infty$

$\displaystyle \lim_{x\to-1^{-}}f(x)=-\infty$

$\displaystyle \lim_{x\to-1^{+}}f(x)=\infty$

$\displaystyle \lim_{x\to\infty}=\infty$

October 8th, 2014, 03:35 PM   #10
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Quote:
 Originally Posted by MarkFL The given function is split at $x=-1$ by a vertical asymptote. To the left is a piece that has a local maximum of -8 and to the right is a piece that has a local minimum of 0. We find: $\displaystyle \lim_{x\to-\infty}=-\infty$ $\displaystyle \lim_{x\to-1^{-}}f(x)=-\infty$ $\displaystyle \lim_{x\to-1^{+}}f(x)=\infty$ $\displaystyle \lim_{x\to\infty}=\infty$
Ummm... Can you post its graph here? I don't know how to.

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