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)
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)
Start with $\displaystyle y = \frac{2x}{3x^2 + 1}$
Differentiate both sides with respect to x:
$\displaystyle \frac{dy}{dx} = \frac{2(3x^2 + 1) - 2x(6x)}{(3x^2 + 1)^2} = \frac{-6x^2 + 2}{(3x^2 + 1)^2}$
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:
$\displaystyle -6x^2 + 2 = 0 \Rightarrow x = \pm \sqrt{\frac{1}{3}}$
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 $\displaystyle \pm \sqrt{\frac{1}{3}}$
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...
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.
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.
Aha, so WolframAlpha says that the maximum value of $\displaystyle \frac{2x}{3x^2 + 1}$ is $\displaystyle \sqrt{\frac{1}{3}}$ and that occurs when $\displaystyle x = \sqrt{\frac{1}{3}}$ Also, the minimum value is $\displaystyle -\sqrt{\frac{1}{3}}$, which occurs when $\displaystyle x = -\sqrt{\frac{1}{3}}$. Calculus worked
There are certain other expressions that you can maximize/minimize without calculus, e.g.
*2nd-degree polynomials, e.g. $\displaystyle 4x^2 + 5x + 1$
*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. $\displaystyle 5 \sin 4x + 2 \cos x$
And others...I doubt there is an easy non-calculus solution to optimizing $\displaystyle y = \frac{2x}{3x^2 + 1} $
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/sec^{2}x
=2/3^{1/2} * sinxcosx
=1/3^{1/2} * sin2x
ie, max = 1/3^{1/2} , min= -1/3^{1/2}
@Johnnylam nice solution.
Basically, let $\displaystyle x = \frac{\tan{\theta}}{\sqrt{3}}$
Then, $\displaystyle y = \frac{2 \tan \theta}{\sqrt{3}(\tan^2 \theta + 1)} = \frac{2 \tan \theta}{\sqrt{3} \sec^2 \theta} = \frac{2}{\sqrt{3}} \sin \theta \cos \theta = \frac{1}{\sqrt{3}} \sin (2\theta)$
Sine achieves its maximum value at 1 and -1, so the max/min values are $\displaystyle \pm \frac{1}{\sqrt{3}}$
Good solution. Idk why I didn't think of a trig substitution...
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.
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