# How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)

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• Jun 14th 2012, 11:56 AM
daigo
How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
I graphed this and I'd like to find the exact values of the maximum/minimum point values. How would I go about doing this? And is it the same method as if it were (2x) / ((3*(x^4)) + 1) or something similar (without infinite y-values obviously)
• Jun 14th 2012, 12:47 PM
richard1234
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
Start with $y = \frac{2x}{3x^2 + 1}$

Differentiate both sides with respect to x:

$\frac{dy}{dx} = \frac{2(3x^2 + 1) - 2x(6x)}{(3x^2 + 1)^2} = \frac{-6x^2 + 2}{(3x^2 + 1)^2}$

The function is continuous and differentiable over the set of real numbers. To find critical points, set the numerator to zero and solve for x:

$-6x^2 + 2 = 0 \Rightarrow x = \pm \sqrt{\frac{1}{3}}$

You can determine if those x-values yield local minimum/maximums y at those two x-values, as well as x-values in the vicinity of $\pm \sqrt{\frac{1}{3}}$
• Jun 14th 2012, 01:43 PM
daigo
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
Sorry, what do you mean by "differentiate?" Is it some operation I haven't learned yet? I'm not sure what you did in that step.
• Jun 14th 2012, 01:46 PM
Reckoner
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
Quote:

Originally Posted by daigo
Sorry, what do you mean by "differentiate?" Is it some operation I haven't learned yet? I'm not sure what you did in that step.

He means find the derivative. Have you taken calculus?
• Jun 14th 2012, 03:26 PM
richard1234
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
Yes, "differentiate" means to find the derivative of a function. Basically, it's like finding the instantaneous rate of change of a function (e.g. the slope of a tangent line at a point on the function). When the derivative of the function is zero, it means that the graph of the function "levels out," that is, the slope of the tangent line is zero.

You probably haven't taken calculus yet, so for now, skip all the quotient rule steps...
• Jun 14th 2012, 03:37 PM
daigo
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
No, I have only taken Algebra so far and I'm still taking it. So I can't do this problem without Calculus?
• Jun 14th 2012, 03:46 PM
Plato
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
Quote:

Originally Posted by daigo
I graphed this and I'd like to find the exact values of the maximum/minimum point values. How would I go about doing this? And is it the same method as if it were (2x) / ((3*(x^4)) + 1) or something similar (without infinite y-values obviously)

From the replies in the tread, it is clear to me that you are not an advanced student.
In this case, use web resources

• Jun 14th 2012, 03:59 PM
Reckoner
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
Quote:

Originally Posted by daigo
No, I have only taken Algebra so far and I'm still taking it. So I can't do this problem without Calculus?

I don't know of another way. We can find the exact maxima and minima of some functions (like parabolas) without calculus, but you'll need derivatives for a more general method.

In the mean time, you can find approximations with a graphing calculator or through numeric methods. Sites like WolframAlpha, which Plato mentioned above, are good resources to use as well. Entering, for example, maximize(2x/(3x^2+1)) will give you the exact maximum.
• Jun 14th 2012, 10:30 PM
richard1234
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
Aha, so WolframAlpha says that the maximum value of $\frac{2x}{3x^2 + 1}$ is $\sqrt{\frac{1}{3}}$ and that occurs when $x = \sqrt{\frac{1}{3}}$ Also, the minimum value is $-\sqrt{\frac{1}{3}}$, which occurs when $x = -\sqrt{\frac{1}{3}}$. Calculus worked :)

There are certain other expressions that you can maximize/minimize without calculus, e.g.

*2nd-degree polynomials, e.g. $4x^2 + 5x + 1$
*A function given a constraint. A classic example of this is maximizing the volume of a rectangular prism given a fixed surface area.
*Functions involving sine or cosine, e.g. $5 \sin 4x + 2 \cos x$

And others...I doubt there is an easy non-calculus solution to optimizing $y = \frac{2x}{3x^2 + 1}$
• Jun 15th 2012, 07:10 AM
johnnylam123
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
Have u learnt tri? I mean sin cos tan?
If so, i'll put x=tany/31/2 {since all real numbers can be represent by tany/31/2}
then it becomes 2(tanx/1/2)/{3(tanx/31/2)2+1}
= {2/31/2 } * tanx/sec2x
=2/31/2 * sinxcosx
=1/31/2 * sin2x

ie, max = 1/31/2 , min= -1/31/2
• Jun 15th 2012, 07:30 AM
richard1234
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
@Johnnylam nice solution.

Basically, let $x = \frac{\tan{\theta}}{\sqrt{3}}$

Then, $y = \frac{2 \tan \theta}{\sqrt{3}(\tan^2 \theta + 1)} = \frac{2 \tan \theta}{\sqrt{3} \sec^2 \theta} = \frac{2}{\sqrt{3}} \sin \theta \cos \theta = \frac{1}{\sqrt{3}} \sin (2\theta)$

Sine achieves its maximum value at 1 and -1, so the max/min values are $\pm \frac{1}{\sqrt{3}}$

Good solution. Idk why I didn't think of a trig substitution...
• Jun 15th 2012, 08:35 AM
Reckoner
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
Very nice indeed, johnnylam123!
• Jun 15th 2012, 08:48 AM
Cbarker1
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
Point-plotting for the function.
2. if x=0, x=-1, x=3.5, x=-2 and x=-3
3. put the values of x into the equation/function
4.plot the points.
• Jun 15th 2012, 11:59 AM
daigo
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
Thank you for that solution. I would've felt so dumb if I went ahead and put "there's no way to get exact values without the graph of it" lol...well, I still feel dumb now for not finding out the answer without help
• Jun 15th 2012, 12:54 PM
richard1234
Re: How to find the maximum/minimum values of (2x) / ((3*(x^2)) + 1)
Well, don't feel bad. It seems like the only "trivial" solutions involve calculus. I probably would've found the trig substitution if I spent more time and effort, but oh well.
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