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

**daigo** · June 14th 2012, 11:56 AM

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)

**richard1234** · June 14th 2012, 12:47 PM

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 $\pm \sqrt{\frac{1}{3}}$

**daigo** · June 14th 2012, 01:43 PM

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.

**Reckoner** · June 14th 2012, 01:46 PM

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?

**richard1234** · June 14th 2012, 03:26 PM

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...

**daigo** · June 14th 2012, 03:37 PM

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

**Plato** · June 14th 2012, 03:46 PM

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

You can see the answer to your question.

**Reckoner** · June 14th 2012, 03:59 PM

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.

**richard1234** · June 14th 2012, 10:30 PM

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 $y = \frac{2x}{3x^2 + 1}$

**johnnylam123** · June 15th 2012, 07:10 AM

Have u learnt tri? I mean sin cos tan?
If so, i'll put x=tany/3^1/2 {since all real numbers can be represent by tany/3^1/2}
then it becomes 2(tanx/^1/2)/{3(tanx/3^1/2)²+1}
= {2/3^1/2 } * tanx/sec²x
=2/3^1/2 * sinxcosx
=1/3^1/2 * sin2x

ie, max = 1/3^1/2 , min= -1/3^1/2

**richard1234** · June 15th 2012, 07:30 AM

@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...

**Reckoner** · June 15th 2012, 08:35 AM

Very nice indeed, johnnylam123!

**Cbarker1** · June 15th 2012, 08:48 AM

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.

**daigo** · June 15th 2012, 11:59 AM

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

**richard1234** · June 15th 2012, 12:54 PM

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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