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We have

$f:R\rightarrow R$


Determine $x$ for which the function has the maximum value. How can I determine the maximum of this function?

Thank you very much in advance!

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I have been teaching calculus for years and I have no idea what you are referring to with $a$ or what you mean by "peak/delta formula". It's got me a little curious! – rschwieb May 26 '12 at 12:29
By a>0 I meant the concavity or convexity of the function. We can determine this for a grade 2 function by only looking at $x^2$ coefficient. By peak/delta I meant $V(\frac{-b}{2a};\frac{-\delta}{4a})$ which would work for a parabola. Didn't exactly know how to translate it into English from my language. Sorry for the misunderstandings! – Grozav Alex Ioan May 26 '12 at 12:33
No problem, I was speculating you meant this. However this is not a quadratic (grade 2) function at all, so it would definitely be inappropriate to apply the method. It's good though that you sought to use a simple method first! Some students just blindly use the sledgehammer... – rschwieb May 26 '12 at 12:36
up vote 1 down vote accepted

You can first observe that it has a horizontal asymptote at y=0, so it is not going to go up or down forever on the ends.

In this case, it's probably best to go straight for the derivative and find the critical points $x=1$ and $x=-1/3$. Doing the first or second derivative test tells you the max occurs at $x=1$, and so that maximum value of $f$ is $1/2$.

Fill in the gaps!

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Thank you for your help! Makes sense :).. The fractions were confusing me, that's why I didn't try the extreme points from the start. – Grozav Alex Ioan May 26 '12 at 12:41

I don't know what you mean by "$a>0$", but you can solve this by just taking the derivative and setting it equal to zero:

$$f'(x) = \frac{(3x^2 + 1) 3 - (3x-1)(6x)}{(3x^2 +1)^2} = \frac{-9x^2 + 6x +3}{(3x^2+1)^2}$$

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

So just find the roots of $3x^2 - 2x - 1=0$, assuming my algebra doesn't have any mistakes. Take whichever one has a larger value when you plug into $f$, and then show that it is, in fact, a maximum.

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