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October 5th, 2014, 07:56 PM   #1
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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?
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October 5th, 2014, 08:26 PM   #2
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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 $\displaystyle y=0$.
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Last edited by MarkFL; October 8th, 2014 at 12:06 AM.
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October 5th, 2014, 09:50 PM   #3
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$$\lim_{x \to -\infty} \frac{2x^2}{x+1} = \lim_{x \to -\infty} \frac{2x}{1+\tfrac1x} = -\infty \\ \lim_{x \to \infty} \frac{2x^2}{x+1} = \lim_{x \to \infty} \frac{2x}{1+\tfrac1x} = \infty$$
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 $x=-1$.
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Last edited by greg1313; October 6th, 2014 at 12:15 AM.
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October 6th, 2014, 07:58 AM   #4
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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.
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October 6th, 2014, 08:21 AM   #5
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Quote:
Originally Posted by Monox D. I-Fly View Post
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
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October 7th, 2014, 11:56 PM   #6
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Quote:
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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?
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October 8th, 2014, 12:07 AM   #7
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Quote:
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The relative maxima is 0 and the relative minima is -8, am I right?
No, switch them.
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October 8th, 2014, 04:13 AM   #8
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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?
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October 8th, 2014, 06:51 AM   #9
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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:

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

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

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

$\displaystyle \lim_{x\to\infty}=\infty$
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October 8th, 2014, 04:35 PM   #10
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Quote:
Originally Posted by MarkFL View Post
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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